Chapter 4: Exploring Algebraic Identities

Expand $$(a + b)^2$$ and compare it with $$a^2 + b^2$$.

$$ (a + b)^2 = a^2 + 2ab + b^2 $$

Take the difference (so that we are comparing the two expressions):

$$ (a + b)^2 - (a^2 + b^2) = a^2 + 2ab + b^2 - (a^2 + b^2) $$

Simplify:

$$ (a + b)^2 - (a^2 + b^2) = 2ab $$

We are told $$ (a + b)^2 < a^2 + b^2 $$. Hence,

$$ 2ab < 0 \[\Rightarrow] ab < 0 $$

The product of two numbers is negative only when the two numbers have opposite signs. Therefore, one of $$a, b$$ is positive and the other is negative (neither is zero, because if one were zero the two expressions would be equal, not unequal).

Again use the expansion:

$$ (a + b)^2 - (a^2 + b^2) = 2ab $$

Now we are told $$ (a + b)^2 > a^2 + b^2 $$. Hence,

$$ 2ab > 0 \[\Rightarrow] ab > 0 $$

The product of two numbers is positive only when they have the same sign. Thus both $$a$$ and $$b$$ are either positive or both are negative (and neither is zero, otherwise the two expressions would be equal).

Step 1 : Expand and compare the two expressions.

Using the identity

$$(a + b)^2 = a^2 + 2ab + b^2,$$

subtract $$a^2 + b^2$$ from both sides to get the difference between the two expressions:

$$(a + b)^2 - (a^2 + b^2) = 2ab.$$

So $$(a + b)^2$$ and $$a^2 + b^2$$ differ by exactly the single term $$2ab$$, and the sign of $$ab$$ determines which side is larger.

Step 2 : Examine all three possible cases for the sign of $$ab$$.

Case 1 — $$ab > 0$$ (both $$a$$ and $$b$$ have the same non-zero sign).
Then $$2ab > 0$$, so

$$(a + b)^2 - (a^2 + b^2) > 0 \quad\Longrightarrow\quad (a + b)^2 > a^2 + b^2.$$

Case 2 — $$ab < 0$$ (one of $$a, b$$ is positive and the other is negative).
Then $$2ab < 0$$, so

$$(a + b)^2 - (a^2 + b^2) < 0 \quad\Longrightarrow\quad (a + b)^2 < a^2 + b^2.$$

For instance, with $$a = 3,\ b = -2$$: $$(a+b)^2 = 1$$ while $$a^2+b^2 = 13$$, confirming $$(a+b)^2 < a^2+b^2$$.

Case 3 — $$ab = 0$$ (at least one of $$a, b$$ is zero).
Then $$2ab = 0$$, so

$$(a + b)^2 - (a^2 + b^2) = 0 \quad\Longrightarrow\quad (a + b)^2 = a^2 + b^2.$$

Step 3 : Identify when equality holds.

From the three cases above, equality between $$(a + b)^2$$ and $$a^2 + b^2$$ occurs precisely when

$$2ab = 0 \quad\Longleftrightarrow\quad ab = 0.$$

The product $$ab$$ is zero exactly when at least one of $$a$$ or $$b$$ is zero (the other may be any real number, including zero).

Identity used: $$(a + b)^2 = a^2 + 2ab + b^2$$

Put $$a = 7x$$ and $$b = 4y$$.

Compute each term:

$$a^2 = (7x)^2 = 49x^2$$
$$2ab = 2 \times 7x \times 4y = 56xy$$
$$b^2 = (4y)^2 = 16y^2$$

Add the three terms:

\[ (7x + 4y)^2 = 49x^2 + 56xy + 16y^2 \]

Identity: $$(a + b)^2 = a^2 + 2ab + b^2$$

Take $$a = \dfrac{7}{5}x$$ and $$b = \dfrac{3}{2}y$$.

Square of the first term:
$$a^2 = \left(\dfrac{7}{5}x\right)^2 = \dfrac{49}{25}x^2$$

Twice the product:
$$2ab = 2 \times \dfrac{7}{5}x \times \dfrac{3}{2}y = \dfrac{21}{5}xy$$

Square of the second term:
$$b^2 = \left(\dfrac{3}{2}y\right)^2 = \dfrac{9}{4}y^2$$

Add them:

\[ \left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2 = \dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2 \]

Identity: $$(a + b)^2 = a^2 + 2ab + b^2$$

Let $$a = 2.5p$$ and $$b = 1.5q$$.

Calculate:

$$a^2 = (2.5p)^2 = 6.25p^2$$
$$2ab = 2 \times 2.5p \times 1.5q = 7.5pq$$
$$b^2 = (1.5q)^2 = 2.25q^2$$

Hence,

\[ (2.5p + 1.5q)^2 = 6.25p^2 + 7.5pq + 2.25q^2 \]

Identity: $$(a + b)^2 = a^2 + 2ab + b^2$$

Choose $$a = \dfrac{3}{4}s$$ and $$b = 8t$$.

Now,

$$a^2 = \left(\dfrac{3}{4}s\right)^2 = \dfrac{9}{16}s^2$$
$$2ab = 2 \times \dfrac{3}{4}s \times 8t = 12st$$
$$b^2 = (8t)^2 = 64t^2$$

Adding,

\[ \left(\dfrac{3}{4}s + 8t\right)^2 = \dfrac{9}{16}s^2 + 12st + 64t^2 \]

Identity: $$(a + b)^2 = a^2 + 2ab + b^2$$

Set $$a = x$$ and $$b = \dfrac{1}{2y}$$.

Compute:

$$a^2 = x^2$$
$$2ab = 2 \times x \times \dfrac{1}{2y} = \dfrac{x}{y}$$
$$b^2 = \left(\dfrac{1}{2y}\right)^2 = \dfrac{1}{4y^2}$$

Therefore,

\[ \left(x + \dfrac{1}{2y}\right)^2 = x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2} \]

Identity: $$(a + b)^2 = a^2 + 2ab + b^2$$

Let $$a = \dfrac{1}{x}$$ and $$b = \dfrac{1}{y}$$.

Then,

$$a^2 = \dfrac{1}{x^2}$$
$$2ab = 2 \times \dfrac{1}{x} \times \dfrac{1}{y} = \dfrac{2}{xy}$$
$$b^2 = \dfrac{1}{y^2}$$

Adding these terms:

\[ \left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2 = \dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2} \]

We use the algebraic identity

$$ (a + b)^2 = a^2 + 2ab + b^2 $$

For $$64^2$$ we split 64 as a convenient tens part and a units part:

$$64 = 60 + 4$$ so here $$a = 60$$ and $$b = 4$$.

Substituting in the identity:

$$64^2 = (60 + 4)^2$$

$$\phantom{64^2}= 60^2 + 2 \times 60 \times 4 + 4^2$$

Calculate each term one by one:

• $$60^2 = 3600$$
• $$2 \times 60 \times 4 = 480$$
• $$4^2 = 16$$

Add the three results:

$$3600 + 480 + 16 = 4096$$

Thus

\[64^2 = 4096\]

Again apply the same identity $$ (a + b)^2 = a^2 + 2ab + b^2 $$.

Write 105 as $$100 + 5$$ so that $$a = 100 ,\; b = 5$$.

$$105^2 = (100 + 5)^2$$

Break it using the identity:

$$= 100^2 + 2 \times 100 \times 5 + 5^2$$

• $$100^2 = 10000$$
• $$2 \times 100 \times 5 = 1000$$
• $$5^2 = 25$$

Add them:

$$10000 + 1000 + 25 = 11025$$

Hence

\[105^2 = 11025\]

Use the same identity $$ (a + b)^2 = a^2 + 2ab + b^2 $$.

Express 205 as $$200 + 5$$ giving $$a = 200 ,\; b = 5$$.

$$205^2 = (200 + 5)^2$$

Applying the identity:

$$= 200^2 + 2 \times 200 \times 5 + 5^2$$

• $$200^2 = 40000$$
• $$2 \times 200 \times 5 = 2000$$
• $$5^2 = 25$$

Add them:

$$40000 + 2000 + 25 = 42025$$

Therefore

\[205^2 = 42025\]

The given trinomial is $$9x^2+24xy+16y^2$$.

Write it in the form $$A^2+2AB+B^2$$.

• First term $$9x^2=(3x)^2\Rightarrow A=3x$$.
• Last term $$16y^2=(4y)^2\Rightarrow B=4y$$.
• Middle term test: $$2AB=2\times(3x)\times(4y)=24xy$$, which matches the given middle term.

Hence the expression is a perfect-square trinomial:

For $$4s^2+20st+25t^2$$, compare with $$A^2+2AB+B^2$$.

• $$4s^2=(2s)^2\Rightarrow A=2s$$,
• $$25t^2=(5t)^2\Rightarrow B=5t$$,
• Middle term: $$2AB=2\times(2s)\times(5t)=20st$$ ✓

Therefore,

Given $$49x^2+28xy+4y^2$$.

• $$49x^2=(7x)^2\Rightarrow A=7x$$,
• $$4y^2=(2y)^2\Rightarrow B=2y$$,
• Test middle term: $$2AB=2\times(7x)\times(2y)=28xy$$ ✓

Thus,

Expression: $$64p^2+\dfrac{32}{3}pq+\dfrac{4}{9}q^2$$.

• $$64p^2=(8p)^2\Rightarrow A=8p$$,
• $$\dfrac{4}{9}q^2=\left(\dfrac{2}{3}q\right)^2\Rightarrow B=\dfrac{2}{3}q$$,
• Middle term check: $$2AB=2\times(8p)\times\left(\dfrac{2}{3}q\right)=\dfrac{32}{3}pq$$ ✓

Hence,

Given $$3a^2+4ab+\dfrac{4}{3}b^2$$.

Step 1 — take out the common factor $$\dfrac{1}{3}$$ to clear fractions:

Step 2 — recognise the perfect square inside the brackets:

• $$(3a)^2=9a^2\;{\small(=A^2)},$$
• $$(2b)^2=4b^2\;{\small(=B^2)},$$
• $$2AB=2\times(3a)\times(2b)=12ab$$ matches the middle term.

Therefore,

Substituting back,

Trinomial: $$\dfrac{9}{5}s^2+6sv+5v^2$$.

Step 1 — factor out $$\dfrac{1}{5}$$:

Step 2 — check the bracketed quadratic:

• $$(3s)^2=9s^2\;{\small(=A^2)},$$
• $$(5v)^2=25v^2\;{\small(=B^2)},$$
• Middle term $$2AB=2\times(3s)\times(5v)=30sv$$ ✓

So,

Hence,

To evaluate $$(79)^2$$ we recognise that 79 is 1 less than 80, an easy number to square.
Let $$a = 80$$ and $$b = 1$$ so that $$79 = a - b$$.
Using the identity $$(a - b)^2 = a^2 - 2ab + b^2$$:

Compute each term step-by-step:

• $$a^2 = 80^2 = 6400$$
• $$2ab = 2 \times 80 \times 1 = 160$$
• $$b^2 = 1^2 = 1$$

Substitute in the identity:

$$79^2 = 6400 - 160 + 1$$

First subtraction: $$6400 - 160 = 6240$$

Final addition: $$6240 + 1 = 6241$$

Hence, $$79^2 = 6241$$.

To evaluate $$(193)^2$$, notice that 193 is 7 less than 200.
Take $$a = 200$$ and $$b = 7$$ so that $$193 = a - b$$.
Apply the identity $$(a - b)^2 = a^2 - 2ab + b^2$$.

Compute each part:

• $$a^2 = 200^2 = 40000$$
• $$2ab = 2 \times 200 \times 7 = 2800$$
• $$b^2 = 7^2 = 49$$

Substitute:

$$193^2 = 40000 - 2800 + 49$$

Simplify in two steps:

First: $$40000 - 2800 = 37200$$

Then: $$37200 + 49 = 37249$$

Therefore, $$193^2 = 37249$$.

To find $$(299)^2$$ we observe that 299 is 1 less than 300.
Choose $$a = 300$$ and $$b = 1$$ so $$299 = a - b$$.
Use the identity $$(a - b)^2 = a^2 - 2ab + b^2$$.

Calculate each expression:

• $$a^2 = 300^2 = 90000$$
• $$2ab = 2 \times 300 \times 1 = 600$$
• $$b^2 = 1^2 = 1$$

Insert these values:

$$299^2 = 90000 - 600 + 1$$

First: $$90000 - 600 = 89400$$

Then: $$89400 + 1 = 89401$$

Thus, $$299^2 = 89401$$.

The convenient benchmark for $$117$$ is $$100$$, so write $$117 = 100 + 17$$.

Apply the identity $$ (a + b)^2 = a^2 + 2ab + b^2 $$ with
$$a = 100, \; b = 17$$.

• $$a^2 = 100^2 = 10000$$
• $$2ab = 2 \times 100 \times 17 = 3400$$
• $$b^2 = 17^2 = 289$$

Add the three terms:

Hence, the value is $$13689$$.

The number $$78$$ is near $$80$$. Write $$78 = 80 - 2$$.

Use the identity $$ (a - b)^2 = a^2 - 2ab + b^2 $$ with
$$a = 80, \; b = 2$$.

• $$a^2 = 80^2 = 6400$$
• $$-2ab = -2 \times 80 \times 2 = -320$$
• $$b^2 = 2^2 = 4$$

Add the terms:

So the square is $$6084$$.

Write $$198 = 200 - 2$$, close to $$200$$.

Apply $$ (a - b)^2 = a^2 - 2ab + b^2 $$ with
$$a = 200, \; b = 2$$.

• $$a^2 = 200^2 = 40000$$
• $$-2ab = -2 \times 200 \times 2 = -800$$
• $$b^2 = 2^2 = 4$$

Adding gives

Thus $$39204$$ is the required square.

Write $$214 = 200 + 14$$.

Use $$ (a + b)^2 = a^2 + 2ab + b^2 $$ with
$$a = 200, \; b = 14$$.

• $$a^2 = 200^2 = 40000$$
• $$2ab = 2 \times 200 \times 14 = 5600$$
• $$b^2 = 14^2 = 196$$

Add them:

Hence the square is $$45796$$.

Express $$1104$$ as $$1100 + 4$$.

Apply $$ (a + b)^2 = a^2 + 2ab + b^2 $$ with
$$a = 1100, \; b = 4$$.

• $$a^2 = 1100^2 = 1210000$$
• $$2ab = 2 \times 1100 \times 4 = 8800$$
• $$b^2 = 4^2 = 16$$

Adding:

Therefore $$1104^2 = 1218816$$.

Write $$1120 = 1100 + 20$$.

Use $$ (a + b)^2 = a^2 + 2ab + b^2 $$ with
$$a = 1100, \; b = 20$$.

• $$a^2 = 1100^2 = 1210000$$
• $$2ab = 2 \times 1100 \times 20 = 44000$$
• $$b^2 = 20^2 = 400$$

Add them:

Thus $$1120^2 = 1254400$$.

The given polynomial is $$16y^2-24y+9$$.

Recognise the pattern $$a^2-2ab+b^2=(a-b)^2$$.

Match each term:

• First term: $$16y^2=(4y)^2$$, so $$a=4y$$.
• Last term: $$9=3^2$$, so $$b=3$$.
• Middle term: $$-24y=-2\times(4y)\times3=-2ab$$ — exactly what the identity requires.

Therefore

The expression is $$\dfrac94s^2+6st+4t^2$$.

Recall the identity $$a^2+2ab+b^2=(a+b)^2$$.

Choose

• $$a=\dfrac{3}{2}s\quad\bigl(a^2=\dfrac{9}{4}s^2\bigr)$$
• $$b=2t\quad\bigl(b^2=4t^2\bigr)$$

Check the middle term:

$$2ab=2\times\dfrac32s\times2t=6st,$$ which matches.

Hence

Write the polynomial term-wise:

$$\dfrac{m^2}{9}+\dfrac{mk}{3}+\dfrac{k^2}{4}+3nk+2mn+9n^2$$

Notice the first three terms form $$\bigl(\dfrac{m}{3}+\dfrac{k}{2}\bigr)^2$$, since

• $$\bigl(\dfrac{m}{3}\bigr)^2=\dfrac{m^2}{9}$$
• $$\bigl(\dfrac{k}{2}\bigr)^2=\dfrac{k^2}{4}$$
• $$2\cdot\dfrac{m}{3}\cdot\dfrac{k}{2}=\dfrac{mk}{3}$$

Next combine all six terms and check whether they come from squaring a trinomial. Let

$$a=\dfrac{m}{3},\;b=\dfrac{k}{2},\;c=3n.$$

Using $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca,$$ we get

• $a^2=\dfrac{m^2}{9}$
• $b^2=\dfrac{k^2}{4}$
• $c^2=9n^2$
• $2ab=\dfrac{mk}{3}$
• $2bc=3nk$
• $2ca=2mn$

These are exactly the six given terms. Therefore

The expression is $$\dfrac{p^2}{16}-2+\dfrac{16}{p^2}$$.

Compare with $$(a-b)^2=a^2-2ab+b^2$$ and choose

$$a=\dfrac{p}{4},\qquad b=\dfrac{4}{p}.$$

Check:

• $a^2=\dfrac{p^2}{16}$
• $b^2=\dfrac{16}{p^2}$
• $-2ab=-2\times\dfrac{p}{4}\times\dfrac{4}{p}=-2$

All three terms match, so

Given polynomial:

$$9a^2+4b^2+c^2-12ab+6ac-4bc$$

Attempt $$(3a-2b+c)^2$$ and expand to verify.

Using $$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$$ with $x=3a,\;y=-2b,\;z=c$:

• $x^2=9a^2$
• $y^2=4b^2$
• $z^2=c^2$
• $2xy=2\times3a\times(-2b)=-12ab$
• $2yz=2\times(-2b)\times c=-4bc$
• $2zx=2\times c\times3a=6ac$

All six terms agree exactly, hence

For the expression $$(p + 3q + 7r)^2$$ we compare it with the identity

$$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca $$

Here we have

• $$a = p$$
• $$b = 3q$$
• $$c = 7r$$

Now compute each required term one by one.

1. Squares of the individual terms:
   $$a^2 = p^2$$
   $$b^2 = (3q)^2 = 9q^2$$
   $$c^2 = (7r)^2 = 49r^2$$
2. Twice each pairwise product:
   $$2ab = 2 \times p \times 3q = 6pq$$
   $$2bc = 2 \times 3q \times 7r = 42qr$$
   $$2ca = 2 \times 7r \times p = 14pr$$

Add all these six terms:

$$p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$$

Thus

For the expression $$(3x - 2y + 4z)^2$$ we again use the identity

$$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca $$

This time choose

• $$a = 3x$$
• $$b = -2y$$
• $$c = 4z$$

Compute each component carefully, keeping the sign of b (that is, $$-2y$$).

1. Squares of the individual terms:
   $$a^2 = (3x)^2 = 9x^2$$
   $$b^2 = (-2y)^2 = 4y^2$$
   $$c^2 = (4z)^2 = 16z^2$$
2. Twice each pairwise product:
   $$2ab = 2 \times (3x) \times (-2y) = -12xy$$
   $$2bc = 2 \times (-2y) \times 4z = -16yz$$
   $$2ca = 2 \times 4z \times 3x = 24xz$$

Combine all terms:

$$9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$$

Hence

We want two numbers whose:

• product is the constant term $$24$$, and
• sum is the coefficient of $$s$$, namely $$-11$$.

The pair $$(-3)$$ and $$(-8)$$ satisfies these conditions because

$$(-3)(-8)=24 \quad\text{and}\quad (-3)+(-8)=-11.$$

Split the middle term:

$$s^2-11s+24=s^2-3s-8s+24.$$

Group and factor:

$$s(s-3)-8(s-3)=(s-3)(s-8).$$

Hence

$$s^2-11s+24=(s-3)(s-8).$$

Let the missing factor be $$Q(x)$$. Then

$$Q(x)\,(x+1) = 3x^2 - 4x - 7.$$

So $$Q(x)$$ is the quotient when $$3x^2 - 4x - 7$$ is divided by $$x+1$$.

Step 1. Divide the leading term: $$3x^2 \div x = 3x.$$

Multiply: $$3x(x+1) = 3x^2 + 3x.$$

Subtract from $$3x^2 - 4x - 7$$, bringing down the constant term $$-7$$:

$$\bigl(3x^2 - 4x - 7\bigr) - \bigl(3x^2 + 3x\bigr) = -7x - 7.$$

Step 2. Divide the new leading term: $$-7x \div x = -7.$$

Multiply: $$-7(x+1) = -7x - 7.$$

Subtract: $$(-7x - 7) - (-7x - 7) = 0.$$ The remainder is $$0$$.

Hence $$Q(x) = 3x - 7$$, and

$$(3x - 7)(x + 1) = 3x^2 - 4x - 7.$$

Check: $$(3x-7)(x+1) = 3x^2 + 3x - 7x - 7 = 3x^2 - 4x - 7.$$ ✔

Assume a factorisation of the form

$$10x^2-11x-6=(2x-a)(bx+2).$$

1. Compare the $$x^2$$ coefficients:

$$2b=10\;\Longrightarrow\; b=5.$$

2. Compare the constant terms:

$$(-a)(2)=-6\;\Longrightarrow\; a=3.$$

3. Check the middle term:

Cross–product coefficient = $$(2)(2x)+(-a)(5x)=4x-5ax.$$ With $$a=3$$ this is $$4x-15x=-11x,$$ matching the given polynomial.

Hence

$$10x^2-11x-6=(2x-3)(5x+2).$$

Multiply the leading and constant coefficients: $$6\times 2=12.$$ Find two numbers whose product is $$12$$ and sum is $$7$$: these are $$3$$ and $$4$$.

Split the middle term:

$$6x^2+7x+2=6x^2+3x+4x+2.$$

Group and factor:

$$3x(2x+1)+2(2x+1)=(3x+2)(2x+1).$$

Therefore

$$6x^2+7x+2=(3x+2)(2x+1).$$

Express 41 as $$40 + 1$$ and use the identity $$(a+b)^2 = a^2 + 2ab + b^2$$.

Take $$a = 40,\; b = 1$$. Then

$$(41)^2 = (40 + 1)^2 = 40^2 + 2 \times 40 \times 1 + 1^2 = 1600 + 80 + 1 = 1681.$$

Write 27 as $$30 - 3$$ and use the identity $$(a-b)^2 = a^2 - 2ab + b^2$$.

Take $$a = 30,\; b = 3$$. Then

$$(27)^2 = (30 - 3)^2 = 30^2 - 2 \times 30 \times 3 + 3^2 = 900 - 180 + 9 = 729.$$

Both numbers are symmetric about $$20$$: $$23 = 20 + 3$$ and $$17 = 20 - 3$$.

Use the identity $$(a+b)(a-b) = a^2 - b^2$$ with $$a = 20,\; b = 3$$:

$$23 \times 17 = (20 + 3)(20 - 3) = 20^2 - 3^2 = 400 - 9 = 391.$$

Write 135 as $$100+35$$ and use $$(a+b)^2=a^2+2ab+b^2$$ with $$a=100,\;b=35$$.

$$135^2=100^2+2\times100\times35+35^2.$$

• $$100^2=10000$$
• $$2\times100\times35=7000$$
• Compute $$35^2$$ via $(30+5)^2$: $$35^2=30^2+2\times30\times5+5^2=900+300+25=1225$$

Adding: $$135^2=10000+7000+1225=18225.$$

Represent 97 as $$100-3$$ and apply $$(a-b)^2=a^2-2ab+b^2$$ with $$a=100,\;b=3$$.

$$97^2=(100-3)^2=100^2-2\times100\times3+3^2=10000-600+9=9409.$$

Convert 29 into $$30-1$$ so that one factor is a multiple of 10.

$$18\times29=18\times(30-1).$$

Use the distributive law $$a(b-c)=ab-ac$$:

$$18\times29=18\times30-18\times1=540-18=522.$$

Take 30 as a convenient base:

$$34=30+4,\;43=30+13.$$

The identity $$(x+a)(x+b)=x^2+x(a+b)+ab$$ gives

$$34\times43=(30+4)(30+13)=30^2+30(4+13)+4\times13.$$

• $$30^2=900$$
• $$30(4+13)=30\times17=510$$
• $$4\times13=52$$

Sum: $$900+510+52=1462.$$

Write 205 as $$200+5$$ and use $$(a+b)^2=a^2+2ab+b^2$$ with $$a=200,\;b=5$$.

$$205^2=(200+5)^2=200^2+2\times200\times5+5^2=40000+2000+25=42025.$$

The given expression is

$$9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc.$$

We look for a perfect-square trinomial of the form

$$(A + B + C)^2 = A^2 + B^2 + C^2 + 2AB + 2BC + 2CA.$$

Match the square terms:

• $$9a^2 = (3a)^2,$$ so choose $$A = 3a.$$
• $$b^2 = (-b)^2,$$ choose $$B = -b$$ (the sign will be confirmed by the cross terms).
• $$4c^2 = (2c)^2,$$ choose $$C = 2c.$$

Now verify the cross terms with $$A = 3a,\; B = -b,\; C = 2c$$:

• $$2AB = 2(3a)(-b) = -6ab$$  ✔
• $$2BC = 2(-b)(2c) = -4bc$$  ✔
• $$2CA = 2(2c)(3a) = 12ac$$  ✔

All six terms match exactly, hence

The expression is $$16s^2 + 25t^2 - 40st.$$

Notice the pattern $$(4s)^2 + ( -5t)^2 + 2(4s)(-5t),$$ which is the expansion of $$ (4s - 5t)^2 $$. Hence,

\[ 16s^2 + 25t^2 - 40st = (4s - 5t)^2. \]

We factor the quadratic $$r^2 - r - 42.$$

Find two numbers whose product is $$-42$$ and whose sum is $$-1$$. These numbers are $$-7$$ and $$6$$.

Split the middle term:

$$r^2 - r - 42 = r^2 - 7r + 6r - 42.$$

Group and factor:

$$r(r - 7) + 6(r - 7) = (r - 7)(r + 6).$$

The expression is $$49g^2 + 14gh + h^2.$$

Recognise the perfect–square pattern:

$$(7g + h)^2 = (7g)^2 + 2(7g)(h) + h^2 = 49g^2 + 14gh + h^2.$$

Therefore,

\[ 49g^2 + 14gh + h^2 = (7g + h)^2. \]

The expression is

$$64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw.$$

Assume it is a perfect square $$(Au + Bv + Cw)^2$$ with $$A=\pm8,\;B=\pm11,\;C=\pm2.$$

Choose signs so that all cross terms match the given ones:

• $$2(8u)(-11v) = -176uv$$ ✔︎
• $$2(8u)(-2w) = -32uw$$ ✔︎
• $$2(-11v)(-2w) = 44vw$$ ✔︎

These are obtained when $$A=8,\;B=-11,\;C=-2.$$ Thus,

$$ (8u - 11v - 2w)^2 = 64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw.$$

Hence,

\[ 64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw = (8u - 11v - 2w)^2. \]

Given expression:

$$\dfrac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}$$

Step 1  Factor the numerator

Take out the common factor $$3$$:

$$3p^2-3pq-18q^2=3\bigl(p^2-pq-6q^2\bigr)$$

Now factor the quadratic inside the bracket:

$$p^2-pq-6q^2=p^2-3pq+2pq-6q^2=(p-3q)(p+2q)$$

Hence

$$3p^2-3pq-18q^2=3(p-3q)(p+2q)$$

Step 2  Factor the denominator

$$p^2+3pq-10q^2=p^2+5pq-2pq-10q^2=(p+5q)(p-2q)$$

Step 3  Write the fraction with factors and cancel

$$\dfrac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}$$

No common non-constant factors are present in numerator and denominator, so no further cancellation is possible (recall the condition $$p^2+3pq-10q^2\neq0$$).

Simplified form:

$$\boxed{\dfrac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}}$$

Given expression:

$$\dfrac{n^3-3n^2m+3nm^2-m^3}{5m^2-10mn+5n^2}$$

Step 1  Recognise the cube expansion in the numerator

$$n^3-3n^2m+3nm^2-m^3=(n-m)^3$$

Step 2  Factor the denominator

Take out the common factor $$5$$:

$$5m^2-10mn+5n^2=5\bigl(m^2-2mn+n^2\bigr)=5(m-n)^2$$

Step 3  Cancel common factors

$$\dfrac{(n-m)^3}{5(n-m)^2}=\dfrac{n-m}{5}$$

Denominator is non-zero because $$n\neq m$$ (given in the question).

Simplified form:

$$\boxed{\dfrac{n-m}{5}}$$

Given expression:

$$\dfrac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}$$

Step 1  Factor the numerator using the identity
Identity: $$a^3+b^3+c^3-3abc=(a+b+c)\bigl(a^2+b^2+c^2-ab-bc-ca\bigr)$$.

Take $$a=w,\;b=x,\;c=-v$$.

Then

$$a^3+b^3+c^3-3abc=w^3+x^3+(-v)^3-3w x(-v)=w^3-v^3+x^3+3wvx$$

Hence

$$w^3-v^3+x^3+3wvx=(w+x-v)\bigl(w^2+x^2+v^2-wx+vx+vw\bigr)$$

Step 2  Factor the denominator

Notice that it matches the square of $$w+x-v$$:

$$\bigl(w+x-v\bigr)^2=w^2+x^2+v^2+2wx-2wv-2vx$$

This is precisely the denominator, so

$$w^2+v^2+x^2-2wv-2vx+2wx=(w+x-v)^2$$

Step 3  Cancel the common factor

$$\dfrac{(w+x-v)\bigl(w^2+x^2+v^2-wx+vx+vw\bigr)}{(w+x-v)^2}=\dfrac{w^2+x^2+v^2-wx+vx+vw}{w+x-v}$$

The denominator is non-zero because $$w+x\neq v$$.

Simplified form:

$$\boxed{\dfrac{w^{2}+v^{2}+x^{2}-wx+wv+vx}{\,w+x-v\,}}$$

Given expression:

$$\dfrac{4y^2-20yz+25z^2}{25z^2-4y^2}$$

Step 1  Factor the numerator

Recognise it as a perfect square:

$$4y^2-20yz+25z^2=(2y-5z)^2$$

Step 2  Factor the denominator

It is a difference of two squares:

$$25z^2-4y^2=(5z)^2-(2y)^2=(5z-2y)(5z+2y)$$

Step 3  Cancel the common factor

Write the denominator as $$-(2y-5z)(5z+2y)$$ to show the common factor clearly:

$$\dfrac{(2y-5z)^2}{-(2y-5z)(5z+2y)}=-\dfrac{2y-5z}{5z+2y}=\dfrac{5z-2y}{5z+2y}$$

The denominator is non-zero because $$5z\pm2y\neq0$$.

Simplified form:

$$\boxed{\dfrac{5z-2y}{5z+2y}}$$

Given expression:

$$\dfrac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$$

Step 1  Factor each quadratic

• $$x^2+x-6=(x+3)(x-2)$$
• $$x^2-7x+12=(x-3)(x-4)$$
• $$x^2-6x+8=(x-2)(x-4)$$
• $$x^2-9=(x-3)(x+3)$$

Step 2  Rewrite the whole fraction with the factors

$$\dfrac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}$$

Step 3  Cancel every common factor

Each factor $$x+3,\;x-2,\;x-3,\;x-4$$ appears once in both numerator and denominator, so every one of them cancels:

$$=1$$

Restrictions: $$x\neq-3,\;2,\;3,\;4$$ (to keep the original denominators non-zero).

Simplified form:

$$\boxed{1}$$

Given expression:

$$\dfrac{p^4-16}{p^2-4p+4}$$

Step 1  Factor the numerator

It is a difference of squares:

$$p^4-16=(p^2)^2-4^2=(p^2-4)(p^2+4)$$

Further factor $$p^2-4$$:

$$p^2-4=(p-2)(p+2)$$

So

$$p^4-16=(p-2)(p+2)(p^2+4)$$

Step 2  Factor the denominator

$$p^2-4p+4=(p-2)^2$$

Step 3  Cancel the common factor $$p-2$$

$$\dfrac{(p-2)(p+2)(p^2+4)}{(p-2)^2}=\dfrac{(p+2)(p^2+4)}{p-2}$$

The denominator is non-zero because $$p\neq2$$.

Simplified form:

$$\boxed{\dfrac{(p+2)(p^2+4)}{p-2}}$$

The expression is $$(-3x + 4)^2$$.
It matches $$(a + b)^2 = a^2 + 2ab + b^2$$ with $$a = -3x$$ and $$b = 4$$.

$$(-3x + 4)^2 = (-3x)^2 + 2(-3x)(4) + 4^2$$
$$ = 9x^2 - 24x + 16$$

Hence,

The factors are $$(2s + 7)(2s - 7)$$.
This is of the form $$(a + b)(a - b) = a^2 - b^2$$ with $$a = 2s$$ and $$b = 7$$.

$$(2s + 7)(2s - 7) = (2s)^2 - 7^2 = 4s^2 - 49$$

Take $$\bigl(p^2 + \dfrac12\bigr)\bigl(p^2 - \dfrac12\bigr)$$.
Again $(a+b)(a-b)=a^2-b^2$ with $$a = p^2$$ and $$b = \dfrac12$$.

$$\left(p^2 + \dfrac12\right)\left(p^2 - \dfrac12\right) = (p^2)^2 - \left(\dfrac12\right)^2 = p^4 - \dfrac14$$

The product is $$(2n + 7)(2n - 7)$$.
Using $(a+b)(a-b)=a^2-b^2\;(a=2n,\;b=7)$:

$$(2n + 7)(2n - 7) = (2n)^2 - 7^2 = 4n^2 - 49$$

We have $$(s - 2t)(s^2 + 2st + 4t^2)$$.
Recall $$(a - b)(a^2 + ab + b^2) = a^3 - b^3$$ (difference of cubes) with $$a = s$$ and $$b = 2t$$.

$$(s - 2t)(s^2 + 2st + 4t^2) = s^3 - (2t)^3 = s^3 - 8t^3$$

Take $$\left(\dfrac{1}{2r} - 4r\right)^2$$.
Use $(a - b)^2 = a^2 - 2ab + b^2$ with $$a = \dfrac{1}{2r},\;b = 4r$$.

$$\left(\dfrac{1}{2r}\right)^2 = \dfrac{1}{4r^2}$$
$$2ab = 2\left(\dfrac{1}{2r}\right)(4r) = 4$$
$$b^2 = (4r)^2 = 16r^2$$
Therefore

$$\left(\dfrac{1}{2r} - 4r\right)^2 = \dfrac{1}{4r^2} - 4 + 16r^2$$

Consider $$(-3m + 4k - l)^2$$.
For three terms $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ with
$$a = -3m,\; b = 4k,\; c = -l$$.

$a^2 = 9m^2$
$b^2 = 16k^2$
$c^2 = l^2$
$2ab = 2(-3m)(4k) = -24mk$
$2bc = 2(4k)(-l) = -8kl$
$2ca = 2(-l)(-3m) = 6lm$

Add them:

$$(-3m + 4k - l)^2 = 9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm$$

The cube is $$\left(x - \dfrac13 y\right)^3$$.
Use $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $$a = x,\; b = \dfrac13 y$$.

$a^3 = x^3$
$-3a^2b = -3x^2\left(\dfrac13 y\right) = -x^2y$
$+3ab^2 = 3x\left(\dfrac13 y\right)^2 = 3x\left(\dfrac{y^2}{9}\right) = \dfrac13 xy^2$
$-b^3 = -\left(\dfrac13 y\right)^3 = -\dfrac{y^3}{27}$

Thus

Compute $$\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3$$ using $(a - b)^3$ with
$$a = \dfrac{7}{2}k,\; b = \dfrac{2}{3}m$$.

$a^3 = \left(\dfrac{7}{2}\right)^3 k^3 = \dfrac{343}{8}k^3$
$-3a^2b$: $a^2 = \left(\dfrac{7}{2}\right)^2k^2 = \dfrac{49}{4}k^2$; then
$-3a^2b = -3\left(\dfrac{49}{4}k^2\right)\left(\dfrac{2}{3}m\right) = -\dfrac{49}{2}k^2m$
$+3ab^2$: $b^2 = \left(\dfrac{2}{3}\right)^2m^2 = \dfrac{4}{9}m^2$; so
$+3ab^2 = 3\left(\dfrac{7}{2}k\right)\left(\dfrac{4}{9}m^2\right) = \dfrac{14}{3}km^2$
$-b^3 = -\left(\dfrac{2}{3}\right)^3m^3 = -\dfrac{8}{27}m^3$

Collecting,

Take the two numbers symmetrically about 19:

$$17=19-2, \; 21=19+2$$

Apply the identity $$\bigl(a-b\bigr)\bigl(a+b\bigr)=a^2-b^2$$ with $$a=19,\; b=2$$:

$$17\times21=(19-2)(19+2)=19^2-2^2$$

Compute each square:

$$19^2=361, \; 2^2=4$$

Subtract:

$$361-4=357$$

Hence

\[17\times21=357\]

Write each number as a deviation from 100:

$$104=100+4, \; 96=100-4$$

Using $$\bigl(a+b\bigr)\bigl(a-b\bigr)=a^2-b^2$$ with $$a=100,\; b=4$$:

$$104\times96=(100+4)(100-4)=100^2-4^2$$

Compute:

$$100^2=10000, \; 4^2=16$$

Hence

$$10000-16=9984$$

\[104\times96=9984\]

Express the numbers around 20:

$$24=20+4, \; 16=20-4$$

Using $$\bigl(a+b\bigr)\bigl(a-b\bigr)=a^2-b^2$$ with $$a=20,\; b=4$$:

$$24\times16=(20+4)(20-4)=20^2-4^2$$

Compute:

$$20^2=400, \; 4^2=16$$

Therefore

$$400-16=384$$

\[24\times16=384\]

Write 147 as $$150-3$$ and use the cube identity $$\bigl(a-b\bigr)^3=a^3-3a^2b+3ab^2-b^3$$ with $$a=150,\; b=3$$:

$$147^3=(150-3)^3$$

Compute each term step by step:

• $$a^3=150^3=3\,375\,000$$
• $$3a^2b=3\times150^2\times3=3\times22\,500\times3=202\,500$$
• $$3ab^2=3\times150\times3^2=3\times150\times9=4\,050$$
• $$b^3=3^3=27$$

Combine with the required signs:

$$147^3=3\,375\,000-202\,500+4\,050-27$$

$$=3\,176\,523$$

\[147^3=3\,176\,523\]

Write 199 as $$200-1$$. Use $$\bigl(a-b\bigr)^3=a^3-3a^2b+3ab^2-b^3$$ with $$a=200,\; b=1$$:

$$199^3=(200-1)^3$$

• $$a^3=200^3=8\,000\,000$$
• $$3a^2b=3\times200^2\times1=3\times40\,000=120\,000$$
• $$3ab^2=3\times200\times1=600$$
• $$b^3=1$$

Combine:

$$199^3=8\,000\,000-120\,000+600-1$$

$$=7\,880\,599$$

\[199^3=7\,880\,599\]

Express 127 as $$125+2$$ and apply $$\bigl(a+b\bigr)^3=a^3+3a^2b+3ab^2+b^3$$ with $$a=125,\; b=2$$:

$$127^3=(125+2)^3$$

• $$a^3=125^3=1\,953\,125$$
• $$3a^2b=3\times125^2\times2=3\times15\,625\times2=93\,750$$
• $$3ab^2=3\times125\times4=1\,500$$
• $$b^3=8$$

Add:

$$127^3=1\,953\,125+93\,750+1\,500+8$$

$$=2\,048\,383$$

\[127^3=2\,048\,383\]

Because the cube of a negative number is negative, compute $$107^3$$ first and then attach a minus sign.

Write $$107=100+7$$. Using $$\bigl(a+b\bigr)^3=a^3+3a^2b+3ab^2+b^3$$ with $$a=100,\; b=7$$:

• $$a^3=100^3=1\,000\,000$$
• $$3a^2b=3\times100^2\times7=3\times10\,000\times7=210\,000$$
• $$3ab^2=3\times100\times49=14\,700$$
• $$b^3=343$$

Sum:

$$107^3=1\,000\,000+210\,000+14\,700+343=1\,225\,043$$

Therefore

$$(-107)^3=-1\,225\,043$$

\[(-107)^3=-1\,225\,043\]

Similarly, $$(-299)^3=-(299^3)$$.

Write $$299=300-1$$ and use $$\bigl(a-b\bigr)^3=a^3-3a^2b+3ab^2-b^3$$ with $$a=300,\; b=1$$:

• $$a^3=300^3=27\,000\,000$$
• $$3a^2b=3\times300^2\times1=3\times90\,000=270\,000$$
• $$3ab^2=3\times300\times1=900$$
• $$b^3=1$$

Combine:

$$299^3=27\,000\,000-270\,000+900-1=26\,730\,899$$

Thus

$$(-299)^3=-26\,730\,899$$

\[(-299)^3=-26\,730\,899\]

Write the first and the last terms as perfect squares:

$$4y^2=(2y)^2\; ,\; \dfrac{1}{16y^2}=\left(\dfrac1{4y}\right)^2$$

Check the middle term of the square $$(2y+\dfrac1{4y})^2$$:

$$2\cdot(2y)\cdot\left(\dfrac1{4y}\right)=1,$$ which is exactly the given middle term. Hence the expression is a perfect square.

\[4y^2+1+\dfrac1{16y^2}=\left(2y+\dfrac1{4y}\right)^2\]

Rewrite each term as a square:

$$9m^2=(3m)^2\; ,\; \dfrac1{25n^2}=\left(\dfrac1{5n}\right)^2$$

The expression is a difference of two squares:

$$a^2-b^2=(a+b)(a-b)$$ with $$a=3m,\; b=\dfrac1{5n}$$

\[9m^2-\dfrac1{25n^2}=\left(3m+\dfrac1{5n}\right)\left(3m-\dfrac1{5n}\right)\]

Recognise each term as a cube:

$$27b^3=(3b)^3\; ,\; \dfrac1{64b^3}=\left(\dfrac1{4b}\right)^3$$

Use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ with $$a=3b,\; b=\dfrac1{4b}$$.

\[27b^3-\dfrac1{64b^3}=\left(3b-\dfrac1{4b}\right)\left[(3b)^2+3b\left(\dfrac1{4b}\right)+\left(\dfrac1{4b}\right)^2\right]\]

Simplify the second factor:

$$(3b)^2=9b^2,\; 3b\left(\dfrac1{4b}\right)=\dfrac34,\; \left(\dfrac1{4b}\right)^2=\dfrac1{16b^2}$$

So

\[27b^3-\dfrac1{64b^3}=\left(3b-\dfrac1{4b}\right)\left(9b^2+\dfrac34+\dfrac1{16b^2}\right)\]

Compare with $$x^2+(a+b)x+ab=(x+a)(x+b)$$.

Need two numbers whose
sum $$=\dfrac56$$ and product $$=\dfrac16$$.

The numbers $$\dfrac12,\;\dfrac13$$ satisfy these conditions.

Split the middle term:

$$x^2+\dfrac56x+\dfrac16=x^2+\dfrac12x+\dfrac13x+\dfrac16$$

Group and factor:

$$x(x+\dfrac12)+\dfrac13(x+\dfrac12)=(x+\dfrac12)(x+\dfrac13)$$

Assume the expression is the cube of a binomial $$ (3u-\dfrac15)^3 $$ and expand to verify.

$$(3u-\dfrac15)^3=(3u)^3-3(3u)^2\left(\dfrac15\right)+3(3u)\left(\dfrac15\right)^2-\left(\dfrac15\right)^3$$

$$=27u^3-\dfrac{27u^2}5+\dfrac{9u}{25}-\dfrac1{125}$$

The result matches the given expression exactly, so

\[27u^3-\dfrac1{125}-\dfrac{27u^2}5+\dfrac{9u}{25}=\left(3u-\dfrac15\right)^3\]

Write both terms as cubes:

$$64y^3=(4y)^3\; ,\; \dfrac1{125}z^3=\left(\dfrac z5\right)^3$$

Use $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ with $$a=4y,\; b=\dfrac z5$$.

\[64y^3+\dfrac1{125}z^3=(4y+\dfrac z5)\bigl(16y^2-\dfrac{4yz}5+\dfrac{z^2}{25}\bigr)\]

Regroup as $$a^3+b^3+c^3-3abc$$ with
$$a=p,\; b=r,\; c=3q\;(\text{since }(3q)^3=27q^3).$$

The identity is
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$

Therefore

\[p^3+27q^3+r^3-9pqr=(p+r+3q)\left(p^2+r^2+9q^2-pr-3qr-3pq\right)\]

Recognise the pattern $$(3m-2)^2=9m^2-12m+4.$$ Since the given expression is identical, it is a perfect square.

\[9m^2-12m+4=(3m-2)^2\]

First factor out $$\dfrac13$$ for convenience:

$$9x^3-\dfrac83y^3+\dfrac{z^3}3+6xyz=\dfrac13\bigl(27x^3-8y^3+z^3+18xyz\bigr).$$

Inside the brackets set
$$a=3x,\; b=z,\; c=-2y.$$

Then $$a^3=27x^3,\; b^3=z^3,\; c^3=-8y^3,$$ and
$$-3abc=-3(3x)(z)(-2y)=+18xyz,$$ giving exactly the inner expression.

Hence it is of the form $$a^3+b^3+c^3-3abc$$.

Apply the identity:

\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).\]

Substitute $$a,b,c$$ and then divide by 3:

\[9x^3-\dfrac83y^3+\dfrac{z^3}3+6xyz=\dfrac13(3x+z-2y)\bigl(9x^2+z^2+4y^2-3xz+2yz+6xy\bigr)\]

We attempt to recognise the expression as a perfect square using the identity

$$(A + B + C)^2 = A^2 + B^2 + C^2 + 2AB + 2BC + 2CA.$$

Step 1 : Read off $$A, B, C$$ from the cross-term coefficients.

Matching $$x, y, z$$ with $$A, B, C$$ respectively, the three cross terms give

• $$2AB = 24\;\Longrightarrow\; AB = 12,$$
• $$2BC = 36\;\Longrightarrow\; BC = 18,$$
• $$2CA = 12\;\Longrightarrow\; CA = 6.$$

Multiplying all three: $$(ABC)^2 = 12 \times 18 \times 6 = 1296,$$ so $$ABC = 36$$ (taking the positive root). Hence

$$A = \dfrac{ABC}{BC} = \dfrac{36}{18} = 2,\quad B = \dfrac{ABC}{CA} = \dfrac{36}{6} = 6,\quad C = \dfrac{ABC}{AB} = \dfrac{36}{12} = 3.$$

This identifies $$A = 2x,\; B = 6y,\; C = 3z.$$

Step 2 : Expand $$(2x + 6y + 3z)^2$$ to verify.

$$(2x + 6y + 3z)^2 = (2x)^2 + (6y)^2 + (3z)^2 + 2(2x)(6y) + 2(6y)(3z) + 2(3z)(2x)$$

$$= 4x^2 + 36y^2 + 9z^2 + 24xy + 36yz + 12xz.$$

Step 3 : Compare with the given expression.

Side by side:

All three cross-term coefficients agree exactly. Only the coefficients of $$y^2$$ and $$z^2$$ appear interchanged in the printed expression: the textbook prints $$9y^2$$ and $$36z^2$$, whereas the perfect-square form requires $$36y^2$$ and $$9z^2.$$ This is a transcription slip in the question (no choice of real numbers $$A, B, C$$ can produce the printed combination consistently). Treating the squared coefficients of $$y$$ and $$z$$ as swapped, the intended factorisation is

Hence, with the typo corrected, the expression factorises as the perfect square $$(2x + 6y + 3z)^2.$$

Assume it is the cube of $$3u-\dfrac16$$ and expand to check:

$$(3u-\dfrac16)^3=(3u)^3-3(3u)^2\left(\dfrac16\right)+3(3u)\left(\dfrac16\right)^2-\left(\dfrac16\right)^3$$

$$=27u^3-\dfrac{27u^2}6+\dfrac{9u}{36}-\dfrac1{216}$$

Simplify the coefficients:

$$-\dfrac{27u^2}6=-\dfrac{9u^2}2,\; \dfrac{9u}{36}=\dfrac u4$$

These match the given terms, so

\[27u^3-\dfrac1{216}-\dfrac{9u^2}2+\dfrac u4=(3u-\dfrac16)^3\]

The given rational expression is $$\dfrac{4x^2+4x+1}{4x^2-1}$$.

1. Factor the numerator.
$$4x^2+4x+1=(2x+1)^2.$$

2. Factor the denominator.
$$4x^2-1=(2x)^2-1^2=(2x-1)(2x+1).$$

3. Substitute the factors.
$$\dfrac{(2x+1)^2}{(2x-1)(2x+1)}.$$

4. Cancel the common factor $$2x+1$$ (which is non-zero by the given condition).
$$\dfrac{2x+1}{2x-1}.$$

Thus the simplified form is $$\dfrac{2x+1}{2x-1}$$, valid for $$x\neq\pm\dfrac12$$.

The expression to be simplified is $$\dfrac{9(3a^3-24b^3)}{9a^2-36b^2}$$.

1. Take out common numerical factors.

$$9(3a^3-24b^3)=27(a^3-8b^3),\qquad 9a^2-36b^2=9(a^2-4b^2).$$

2. Factor by standard identities.

• Difference of cubes: $$a^3-(2b)^3=(a-2b)(a^2+2ab+4b^2).$$
• Difference of squares: $$a^2-4b^2=(a-2b)(a+2b).$$

So

$$27(a^3-8b^3)=27(a-2b)(a^2+2ab+4b^2),$$
$$9(a^2-4b^2)=9(a-2b)(a+2b).$$

3. Rewrite the fraction.

$$\dfrac{27(a-2b)(a^2+2ab+4b^2)}{9(a-2b)(a+2b)}.$$

4. Cancel the common factors. Since $$a\neq2b$$, divide numerator and denominator by $$9(a-2b)$$:

$$\dfrac{3(a^2+2ab+4b^2)}{a+2b}.$$

Hence the simplified form is $$\dfrac{3(a^2+2ab+4b^2)}{a+2b}$$, with the restrictions $$a\neq2b,\;a\neq-2b$$.

We have to simplify $$\dfrac{s^3+125t^3}{s^2-2st-35t^2}$$.

1. Factor the numerator (sum of cubes).
$$s^3+125t^3=s^3+(5t)^3=(s+5t)(s^2-5st+25t^2).$$

2. Factor the denominator (quadratic in $$s$$).
Find two numbers whose product is $$-35$$ and whose sum is $$-2$$; they are $$5$$ and $$-7$$. Hence

$$s^2-2st-35t^2=(s-7t)(s+5t).$$

3. Express the fraction in factored form.

$$\dfrac{(s+5t)(s^2-5st+25t^2)}{(s-7t)(s+5t)}.$$

4. Cancel the common factor $$s+5t$$ (with $$s\neq-5t$$).
$$\dfrac{s^2-5st+25t^2}{s-7t}.$$

Therefore the simplified expression is $$\dfrac{s^2-5st+25t^2}{s-7t}$$, provided $$s\neq-5t$$ and $$s\neq7t$$.

We compare the given area expression with the standard algebraic identity

• $$x^2 - 2xy + y^2 = (x - y)^2$$  or equivalently  $$x^2 - 2xy + y^2 = (x - y)(x - y)$$

Write the coefficients of the given trinomial $$25a^2 - 30ab + 9b^2$$ in a suggestive squared form.

1. Observe the first and last terms:
   • $$25a^2 = (5a)^2$$
   • $$9b^2 = (3b)^2$$
2. Now check the middle term:
   • For $$(5a)^2 - 2(5a)(3b) + (3b)^2$$ we would get the middle term $$-2 \times 5a \times 3b = -30ab$$, which matches exactly.

Hence the expression fits the identity with $$x = 5a$$ and $$y = 3b$$:

Therefore, if a rectangle has this area, one convenient pair of dimensions is

• Length $$= 5a - 3b$$ units,
• Breadth $$= 5a - 3b$$ units.

(Any arrangement that simply swaps “length” and “breadth” is equally acceptable because multiplying in the opposite order does not change the area.)

The given area expression is $$36s^2 - 49t^2$$. It is a ‘difference of two squares’, so we recall the identity

• $$x^2 - y^2 = (x + y)(x - y)$$

Rewrite each square explicitly:

• $$36s^2 = (6s)^2$$
• $$49t^2 = (7t)^2$$

Apply the identity with $$x = 6s$$ and $$y = 7t$$:

Thus, suitable dimensions for the rectangle are

• Length $$= 6s + 7t$$ units,
• Breadth $$= 6s - 7t$$ units.

(If it is required that the numerical value be positive, we must ensure $$6s \ge 7t$$ so that $$6s - 7t \ge 0$$, or simply choose the absolute values.)

The volume of the cuboid is given by $$6a^2-24b^2$$.

Step 1 – Taking out the H.C.F.

The highest common factor of the two terms is $$6$$, so

$$6a^2-24b^2 = 6(a^2-4b^2).$$

Step 2 – Difference of squares

$$a^2-4b^2$$ can be written as $$a^2-(2b)^2$$, i.e. a difference of two squares:

$$a^2-4b^2=(a-2b)(a+2b).$$

Hence

Since the volume of a cuboid equals $$\text{length}\times\text{breadth}\times\text{height}$$, we may take

• Length $$=6$$,
• Breadth $$=a-2b$$,
• Height $$=a+2b$$.

(Any permutation of these three factors is equally valid.)

The volume expression is $$3ps^2-15ps+12p$$.

Step 1 – Taking out the common factor

Every term contains $$3p$$, hence

$$3ps^2-15ps+12p = 3p(s^2-5s+4).$$

Step 2 – Factorising the quadratic

For $$s^2-5s+4$$, we need two numbers whose product is $$+4$$ and whose sum is $$-5$$. These numbers are $$-1$$ and $$-4$$.

Therefore

$$s^2-5s+4 = (s-1)(s-4).$$

Substituting back,

Thus the three factors can represent the three linear dimensions of a cuboid. Choose

• Length $$=3p$$,
• Breadth $$=s-1$$,
• Height $$=s-4$$.

(Any rearrangement of the three factors is acceptable.)

Given $$x+y=-4$$.

Recall the identity $$x^3+y^3=(x+y)^3-3xy(x+y).$$

Let the required value be $$E$$.

$$E=x^3+y^3-12xy+64$$
$$\;\;\;=(x+y)^3-3xy(x+y)-12xy+64$$ (using the identity)

Substitute $$x+y=-4$$:

$$E=(-4)^3-3xy(-4)-12xy+64=-64+12xy-12xy+64.$$

The $$xy$$ terms cancel and $$-64+64=0$$.

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Given $$x=2y+6\;\Rightarrow\;x-2y=6.$$

Write the first two terms as a difference of cubes:

$$x^3-8y^3=(x-2y)(x^2+2xy+4y^2).$$

Since $$x-2y=6$$,

$$x^3-8y^3=6(x^2+2xy+4y^2).$$

Now the entire expression $$F$$ is

$$F=x^3-8y^3-36xy-216$$
$$\;\;\;=6(x^2+2xy+4y^2)-36xy-216.$$

Factor out 6:

$$F=6\bigl(x^2+2xy+4y^2-6xy-36\bigr).$$

Simplify inside the bracket:

$$x^2+2xy+4y^2-6xy-36=x^2-4xy+4y^2-36=(x-2y)^2-36.$$

But $$x-2y=6\;\Rightarrow\;(x-2y)^2=36$$, hence

$$(x-2y)^2-36=36-36=0.$$

Therefore $$F=6\times0=0.$$

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