Chapter 7: Binomial Theorem

By the Binomial Theorem, $$(p+q)^n = \sum_{k=0}^{n}\binom{n}{k}\,p^{n-k}q^k$$. Here we take $$p = x^2$$, $$q = \dfrac{3}{x}$$ and $$n = 4$$.

$$\left(x^2 + \dfrac{3}{x}\right)^4 = \binom{4}{0}(x^2)^4 + \binom{4}{1}(x^2)^3\left(\dfrac{3}{x}\right) + \binom{4}{2}(x^2)^2\left(\dfrac{3}{x}\right)^2 + \binom{4}{3}(x^2)\left(\dfrac{3}{x}\right)^3 + \binom{4}{4}\left(\dfrac{3}{x}\right)^4$$

The required binomial coefficients are $$\binom{4}{0}=1,\ \binom{4}{1}=4,\ \binom{4}{2}=6,\ \binom{4}{3}=4,\ \binom{4}{4}=1$$.

Now evaluate each term separately:

$$\binom{4}{0}(x^2)^4 = 1\cdot x^8 = x^8$$

$$\binom{4}{1}(x^2)^3\left(\dfrac{3}{x}\right) = 4\cdot x^6\cdot\dfrac{3}{x} = 12x^5$$

$$\binom{4}{2}(x^2)^2\left(\dfrac{3}{x}\right)^2 = 6\cdot x^4\cdot\dfrac{9}{x^2} = 54x^2$$

$$\binom{4}{3}(x^2)\left(\dfrac{3}{x}\right)^3 = 4\cdot x^2\cdot\dfrac{27}{x^3} = \dfrac{108}{x}$$

$$\binom{4}{4}\left(\dfrac{3}{x}\right)^4 = 1\cdot\dfrac{81}{x^4} = \dfrac{81}{x^4}$$

Adding all the terms gives the expansion:

$$\left(x^2 + \dfrac{3}{x}\right)^4 = x^8 + 12x^5 + 54x^2 + \dfrac{108}{x} + \dfrac{81}{x^4}$$

We express $$98$$ as the difference of two numbers whose powers are easy to handle. Write $$98 = 100 - 2$$.

$$(98)^5 = (100-2)^5$$

Applying the Binomial Theorem with $$a = 100$$, $$b = 2$$ and $$n = 5$$:

$$(100-2)^5 = \binom{5}{0}(100)^5 - \binom{5}{1}(100)^4(2) + \binom{5}{2}(100)^3(2)^2 - \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)(2)^4 - \binom{5}{5}(2)^5$$

Using $$\binom{5}{0}=1,\ \binom{5}{1}=5,\ \binom{5}{2}=10,\ \binom{5}{3}=10,\ \binom{5}{4}=5,\ \binom{5}{5}=1$$:

$$= 10000000000 - 5(100000000)(2) + 10(1000000)(4) - 10(10000)(8) + 5(100)(16) - 32$$

$$= 10000000000 - 1000000000 + 40000000 - 800000 + 8000 - 32$$

Adding the positive terms and the negative terms separately:

$$= (10000000000 + 40000000 + 8000) - (1000000000 + 800000 + 32)$$

$$= 10040008000 - 1000800032 = 9039207968$$

Hence $$(98)^5 = 9039207968$$.

We split $$1.01$$ as $$1 + 0.01$$ so that the Binomial Theorem can be applied.

$$(1.01)^{1000000} = (1 + 0.01)^{1000000}$$

Writing the first two terms of the expansion explicitly:

$$(1+0.01)^{1000000} = \binom{1000000}{0} + \binom{1000000}{1}(0.01) + \binom{1000000}{2}(0.01)^2 + \cdots$$

$$= 1 + 1000000 \times 0.01 + (\text{other positive terms})$$

$$= 1 + 10000 + (\text{other positive terms})$$

Every remaining term of the expansion is a product of positive quantities, hence positive. Dropping them can only decrease the value, so

$$(1.01)^{1000000} > 1 + 10000 > 10000$$

Therefore $$(1.01)^{1000000}$$ is larger than $$10{,}000$$.

For two numbers $$a$$ and $$b$$, if we can write $$a = bq + r$$, then $$b$$ divides $$a$$ with quotient $$q$$ and remainder $$r$$. So to show $$6^n - 5n$$ leaves remainder $$1$$ on division by $$25$$, we must prove that $$6^n - 5n = 25k + 1$$ for some natural number $$k$$.

By the Binomial Theorem, $$(1+a)^n = \binom{n}{0} + \binom{n}{1}a + \binom{n}{2}a^2 + \cdots + \binom{n}{n}a^n$$.

Put $$a = 5$$:

$$(1+5)^n = \binom{n}{0} + \binom{n}{1}5 + \binom{n}{2}5^2 + \binom{n}{3}5^3 + \cdots + \binom{n}{n}5^n$$

That is,

$$6^n = 1 + 5n + 5^2\binom{n}{2} + 5^3\binom{n}{3} + \cdots + 5^n$$

Subtracting $$5n$$ from both sides:

$$6^n - 5n = 1 + 5^2\binom{n}{2} + 5^3\binom{n}{3} + \cdots + 5^n$$

Every term after the $$1$$ contains a factor $$5^2 = 25$$, so take $$25$$ common from them:

$$6^n - 5n = 1 + 25\left(\binom{n}{2} + 5\binom{n}{3} + \cdots + 5^{n-2}\right)$$

Writing $$k = \binom{n}{2} + 5\binom{n}{3} + \cdots + 5^{n-2}$$, which is a natural number, we get

$$6^n - 5n = 25k + 1$$

This shows that $$6^n - 5n$$ leaves remainder $$1$$ when divided by $$25$$.

By the Binomial Theorem, $$(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$$. Take $$a = 1$$, $$b = -2x$$ and $$n = 5$$.

The row of binomial coefficients for index $$5$$ is $$1,\ 5,\ 10,\ 10,\ 5,\ 1$$.

$$(1-2x)^5 = \binom{5}{0}(1)^5 + \binom{5}{1}(1)^4(-2x) + \binom{5}{2}(1)^3(-2x)^2 + \binom{5}{3}(1)^2(-2x)^3 + \binom{5}{4}(1)(-2x)^4 + \binom{5}{5}(-2x)^5$$

Evaluating each term:

$$= 1 + 5(-2x) + 10(4x^2) + 10(-8x^3) + 5(16x^4) + (-32x^5)$$

$$= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$$

Hence $$(1-2x)^5 = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$$.

By the Binomial Theorem, take $$a = \dfrac{2}{x}$$, $$b = -\dfrac{x}{2}$$ and $$n = 5$$. The coefficients for index $$5$$ are $$1,\ 5,\ 10,\ 10,\ 5,\ 1$$.

$$\left(\dfrac{2}{x} - \dfrac{x}{2}\right)^5 = \binom{5}{0}\left(\dfrac{2}{x}\right)^5 + \binom{5}{1}\left(\dfrac{2}{x}\right)^4\left(-\dfrac{x}{2}\right) + \binom{5}{2}\left(\dfrac{2}{x}\right)^3\left(-\dfrac{x}{2}\right)^2 + \binom{5}{3}\left(\dfrac{2}{x}\right)^2\left(-\dfrac{x}{2}\right)^3 + \binom{5}{4}\left(\dfrac{2}{x}\right)\left(-\dfrac{x}{2}\right)^4 + \binom{5}{5}\left(-\dfrac{x}{2}\right)^5$$

Evaluating each term:

$$\binom{5}{0}\left(\dfrac{2}{x}\right)^5 = \dfrac{32}{x^5}$$

$$\binom{5}{1}\left(\dfrac{2}{x}\right)^4\left(-\dfrac{x}{2}\right) = 5\cdot\dfrac{16}{x^4}\cdot\left(-\dfrac{x}{2}\right) = -\dfrac{40}{x^3}$$

$$\binom{5}{2}\left(\dfrac{2}{x}\right)^3\left(-\dfrac{x}{2}\right)^2 = 10\cdot\dfrac{8}{x^3}\cdot\dfrac{x^2}{4} = \dfrac{20}{x}$$

$$\binom{5}{3}\left(\dfrac{2}{x}\right)^2\left(-\dfrac{x}{2}\right)^3 = 10\cdot\dfrac{4}{x^2}\cdot\left(-\dfrac{x^3}{8}\right) = -5x$$

$$\binom{5}{4}\left(\dfrac{2}{x}\right)\left(-\dfrac{x}{2}\right)^4 = 5\cdot\dfrac{2}{x}\cdot\dfrac{x^4}{16} = \dfrac{5x^3}{8}$$

$$\binom{5}{5}\left(-\dfrac{x}{2}\right)^5 = -\dfrac{x^5}{32}$$

Adding all the terms:

$$\left(\dfrac{2}{x} - \dfrac{x}{2}\right)^5 = \dfrac{32}{x^5} - \dfrac{40}{x^3} + \dfrac{20}{x} - 5x + \dfrac{5x^3}{8} - \dfrac{x^5}{32}$$

By the Binomial Theorem, take $$a = 2x$$, $$b = -3$$ and $$n = 6$$. The coefficients for index $$6$$ are $$1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$$.

$$(2x-3)^6 = \binom{6}{0}(2x)^6 + \binom{6}{1}(2x)^5(-3) + \binom{6}{2}(2x)^4(-3)^2 + \binom{6}{3}(2x)^3(-3)^3 + \binom{6}{4}(2x)^2(-3)^4 + \binom{6}{5}(2x)(-3)^5 + \binom{6}{6}(-3)^6$$

Evaluating each term:

$$\binom{6}{0}(2x)^6 = 64x^6$$

$$\binom{6}{1}(2x)^5(-3) = 6(32x^5)(-3) = -576x^5$$

$$\binom{6}{2}(2x)^4(-3)^2 = 15(16x^4)(9) = 2160x^4$$

$$\binom{6}{3}(2x)^3(-3)^3 = 20(8x^3)(-27) = -4320x^3$$

$$\binom{6}{4}(2x)^2(-3)^4 = 15(4x^2)(81) = 4860x^2$$

$$\binom{6}{5}(2x)(-3)^5 = 6(2x)(-243) = -2916x$$

$$\binom{6}{6}(-3)^6 = 729$$

Adding all the terms:

$$(2x-3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$$

By the Binomial Theorem, take $$a = \dfrac{x}{3}$$, $$b = \dfrac{1}{x}$$ and $$n = 5$$. The coefficients for index $$5$$ are $$1,\ 5,\ 10,\ 10,\ 5,\ 1$$.

$$\left(\dfrac{x}{3} + \dfrac{1}{x}\right)^5 = \binom{5}{0}\left(\dfrac{x}{3}\right)^5 + \binom{5}{1}\left(\dfrac{x}{3}\right)^4\left(\dfrac{1}{x}\right) + \binom{5}{2}\left(\dfrac{x}{3}\right)^3\left(\dfrac{1}{x}\right)^2 + \binom{5}{3}\left(\dfrac{x}{3}\right)^2\left(\dfrac{1}{x}\right)^3 + \binom{5}{4}\left(\dfrac{x}{3}\right)\left(\dfrac{1}{x}\right)^4 + \binom{5}{5}\left(\dfrac{1}{x}\right)^5$$

Evaluating each term:

$$\binom{5}{0}\left(\dfrac{x}{3}\right)^5 = \dfrac{x^5}{243}$$

$$\binom{5}{1}\left(\dfrac{x}{3}\right)^4\left(\dfrac{1}{x}\right) = 5\cdot\dfrac{x^4}{81}\cdot\dfrac{1}{x} = \dfrac{5x^3}{81}$$

$$\binom{5}{2}\left(\dfrac{x}{3}\right)^3\left(\dfrac{1}{x}\right)^2 = 10\cdot\dfrac{x^3}{27}\cdot\dfrac{1}{x^2} = \dfrac{10x}{27}$$

$$\binom{5}{3}\left(\dfrac{x}{3}\right)^2\left(\dfrac{1}{x}\right)^3 = 10\cdot\dfrac{x^2}{9}\cdot\dfrac{1}{x^3} = \dfrac{10}{9x}$$

$$\binom{5}{4}\left(\dfrac{x}{3}\right)\left(\dfrac{1}{x}\right)^4 = 5\cdot\dfrac{x}{3}\cdot\dfrac{1}{x^4} = \dfrac{5}{3x^3}$$

$$\binom{5}{5}\left(\dfrac{1}{x}\right)^5 = \dfrac{1}{x^5}$$

Adding all the terms:

$$\left(\dfrac{x}{3} + \dfrac{1}{x}\right)^5 = \dfrac{x^5}{243} + \dfrac{5x^3}{81} + \dfrac{10x}{27} + \dfrac{10}{9x} + \dfrac{5}{3x^3} + \dfrac{1}{x^5}$$

By the Binomial Theorem, take $$a = x$$, $$b = \dfrac{1}{x}$$ and $$n = 6$$. The coefficients for index $$6$$ are $$1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$$.

$$\left(x + \dfrac{1}{x}\right)^6 = \binom{6}{0}x^6 + \binom{6}{1}x^5\left(\dfrac{1}{x}\right) + \binom{6}{2}x^4\left(\dfrac{1}{x}\right)^2 + \binom{6}{3}x^3\left(\dfrac{1}{x}\right)^3 + \binom{6}{4}x^2\left(\dfrac{1}{x}\right)^4 + \binom{6}{5}x\left(\dfrac{1}{x}\right)^5 + \binom{6}{6}\left(\dfrac{1}{x}\right)^6$$

Evaluating each term:

$$\binom{6}{0}x^6 = x^6$$

$$\binom{6}{1}x^5\left(\dfrac{1}{x}\right) = 6x^4$$

$$\binom{6}{2}x^4\left(\dfrac{1}{x}\right)^2 = 15x^2$$

$$\binom{6}{3}x^3\left(\dfrac{1}{x}\right)^3 = 20$$

$$\binom{6}{4}x^2\left(\dfrac{1}{x}\right)^4 = \dfrac{15}{x^2}$$

$$\binom{6}{5}x\left(\dfrac{1}{x}\right)^5 = \dfrac{6}{x^4}$$

$$\binom{6}{6}\left(\dfrac{1}{x}\right)^6 = \dfrac{1}{x^6}$$

Adding all the terms:

$$\left(x + \dfrac{1}{x}\right)^6 = x^6 + 6x^4 + 15x^2 + 20 + \dfrac{15}{x^2} + \dfrac{6}{x^4} + \dfrac{1}{x^6}$$

We write $$96$$ close to a round number: $$96 = 100 - 4$$.

$$(96)^3 = (100-4)^3$$

Applying the Binomial Theorem with $$a = 100$$, $$b = 4$$, $$n = 3$$:

$$(100-4)^3 = \binom{3}{0}(100)^3 - \binom{3}{1}(100)^2(4) + \binom{3}{2}(100)(4)^2 - \binom{3}{3}(4)^3$$

$$= (100)^3 - 3(100)^2(4) + 3(100)(16) - 64$$

$$= 1000000 - 120000 + 4800 - 64$$

$$= 884736$$

Hence $$(96)^3 = 884736$$.

We write $$102 = 100 + 2$$.

$$(102)^5 = (100+2)^5$$

Applying the Binomial Theorem with $$a = 100$$, $$b = 2$$, $$n = 5$$ (coefficients $$1, 5, 10, 10, 5, 1$$):

$$(100+2)^5 = \binom{5}{0}(100)^5 + \binom{5}{1}(100)^4(2) + \binom{5}{2}(100)^3(2)^2 + \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)(2)^4 + \binom{5}{5}(2)^5$$

$$= 10000000000 + 5(100000000)(2) + 10(1000000)(4) + 10(10000)(8) + 5(100)(16) + 32$$

$$= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32$$

Adding all the terms:

$$= 11040808032$$

Hence $$(102)^5 = 11040808032$$.

We write $$101 = 100 + 1$$.

$$(101)^4 = (100+1)^4$$

Applying the Binomial Theorem with $$a = 100$$, $$b = 1$$, $$n = 4$$ (coefficients $$1, 4, 6, 4, 1$$):

$$(100+1)^4 = \binom{4}{0}(100)^4 + \binom{4}{1}(100)^3(1) + \binom{4}{2}(100)^2(1)^2 + \binom{4}{3}(100)(1)^3 + \binom{4}{4}(1)^4$$

$$= (100)^4 + 4(100)^3 + 6(100)^2 + 4(100) + 1$$

$$= 100000000 + 4000000 + 60000 + 400 + 1$$

$$= 104060401$$

Hence $$(101)^4 = 104060401$$.

We write $$99 = 100 - 1$$.

$$(99)^5 = (100-1)^5$$

Applying the Binomial Theorem with $$a = 100$$, $$b = 1$$, $$n = 5$$ (coefficients $$1, 5, 10, 10, 5, 1$$):

$$(100-1)^5 = \binom{5}{0}(100)^5 - \binom{5}{1}(100)^4(1) + \binom{5}{2}(100)^3(1)^2 - \binom{5}{3}(100)^2(1)^3 + \binom{5}{4}(100)(1)^4 - \binom{5}{5}(1)^5$$

$$= (100)^5 - 5(100)^4 + 10(100)^3 - 10(100)^2 + 5(100) - 1$$

$$= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1$$

Adding the positive and negative terms separately:

$$= (10000000000 + 10000000 + 500) - (500000000 + 100000 + 1)$$

$$= 10010000500 - 500100001 = 9509900499$$

Hence $$(99)^5 = 9509900499$$.

We split $$1.1$$ as $$1 + 0.1$$ and apply the Binomial Theorem.

$$(1.1)^{10000} = (1 + 0.1)^{10000}$$

Writing the first two terms of the expansion explicitly:

$$(1+0.1)^{10000} = \binom{10000}{0} + \binom{10000}{1}(0.1) + \binom{10000}{2}(0.1)^2 + \cdots$$

$$= 1 + 10000 \times 0.1 + (\text{other positive terms})$$

$$= 1 + 1000 + (\text{other positive terms})$$

Every remaining term of the expansion is positive, so the full sum exceeds $$1 + 1000$$:

$$(1.1)^{10000} > 1 + 1000 > 1000$$

Therefore $$(1.1)^{10000}$$ is larger than $$1000$$.

Expand both fourth powers by the Binomial Theorem (coefficients for index $$4$$ are $$1, 4, 6, 4, 1$$):

$$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$

$$(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$$

Subtracting, the terms $$a^4$$, $$6a^2b^2$$ and $$b^4$$ cancel, while the odd-power terms double:

$$(a+b)^4 - (a-b)^4 = 8a^3b + 8ab^3$$

$$(a+b)^4 - (a-b)^4 = 8ab(a^2 + b^2)$$

Now put $$a = \sqrt{3}$$ and $$b = \sqrt{2}$$, so $$a^2 = 3$$, $$b^2 = 2$$ and $$ab = \sqrt{3}\cdot\sqrt{2} = \sqrt{6}$$:

$$(\sqrt{3}+\sqrt{2})^4 - (\sqrt{3}-\sqrt{2})^4 = 8\sqrt{6}\,(3 + 2) = 8\sqrt{6}\times 5 = 40\sqrt{6}$$

Expand both sixth powers by the Binomial Theorem (coefficients for index $$6$$ are $$1, 6, 15, 20, 15, 6, 1$$):

$$(x+1)^6 = x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1$$

$$(x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1$$

Adding, the odd-power terms cancel and the even-power terms double:

$$(x+1)^6 + (x-1)^6 = 2\left(x^6 + 15x^4 + 15x^2 + 1\right)$$

Now put $$x = \sqrt{2}$$, so $$x^2 = 2$$, $$x^4 = 4$$ and $$x^6 = 8$$:

$$(\sqrt{2}+1)^6 + (\sqrt{2}-1)^6 = 2\left(8 + 15(4) + 15(2) + 1\right)$$

$$= 2\left(8 + 60 + 30 + 1\right) = 2(99) = 198$$

We must show $$9^{n+1} - 8n - 9$$ is a multiple of $$64$$. Write $$9 = 1 + 8$$ and expand $$9^{n+1}$$ by the Binomial Theorem.

$$9^{n+1} = (1+8)^{n+1} = \sum_{k=0}^{n+1}\binom{n+1}{k}8^k$$

$$= \binom{n+1}{0} + \binom{n+1}{1}8 + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \cdots + \binom{n+1}{n+1}8^{n+1}$$

The first two terms are $$\binom{n+1}{0} = 1$$ and $$\binom{n+1}{1}8 = 8(n+1) = 8n + 8$$. So

$$9^{n+1} = 1 + (8n + 8) + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \cdots$$

$$9^{n+1} = 9 + 8n + 8^2\left(\binom{n+1}{2} + \binom{n+1}{3}8 + \cdots\right)$$

Subtracting $$8n$$ and $$9$$ from both sides:

$$9^{n+1} - 8n - 9 = 8^2\left(\binom{n+1}{2} + \binom{n+1}{3}8 + \cdots\right)$$

$$9^{n+1} - 8n - 9 = 64\left(\binom{n+1}{2} + \binom{n+1}{3}8 + \cdots\right)$$

The bracket is an integer, so $$9^{n+1} - 8n - 9$$ is $$64$$ times an integer. Hence it is divisible by $$64$$ for every positive integer $$n$$.

Recall the special case of the Binomial Theorem with $$a = 1$$:

$$(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n = \sum_{r=0}^{n}\binom{n}{r}x^r$$

This identity holds for every value of $$x$$. Substitute $$x = 3$$:

$$(1 + 3)^n = \sum_{r=0}^{n}\binom{n}{r}\,3^r$$

The left side is $$(1+3)^n = 4^n$$, therefore

$$\sum_{r=0}^{n} 3^r \, {}^{n}C_r = 4^n$$

which is the required result.

Following the hint, write $$a = (a-b) + b$$ and raise both sides to the power $$n$$:

$$a^n = \big((a-b) + b\big)^n$$

Expand the right side by the Binomial Theorem, treating $$(a-b)$$ as the first quantity and $$b$$ as the second:

$$a^n = \binom{n}{0}(a-b)^0 b^n + \binom{n}{1}(a-b)^1 b^{n-1} + \binom{n}{2}(a-b)^2 b^{n-2} + \cdots + \binom{n}{n}(a-b)^n$$

The very first term ($$k=0$$) is $$\binom{n}{0}b^n = b^n$$. Hence

$$a^n = b^n + \binom{n}{1}(a-b)b^{n-1} + \binom{n}{2}(a-b)^2 b^{n-2} + \cdots + (a-b)^n$$

Transposing $$b^n$$ to the left:

$$a^n - b^n = \binom{n}{1}(a-b)b^{n-1} + \binom{n}{2}(a-b)^2 b^{n-2} + \cdots + (a-b)^n$$

Every term on the right contains at least one factor of $$(a-b)$$, so we can take $$(a-b)$$ common:

$$a^n - b^n = (a-b)\left[\binom{n}{1}b^{n-1} + \binom{n}{2}(a-b)b^{n-2} + \cdots + (a-b)^{n-1}\right]$$

Since $$a$$ and $$b$$ are integers, the expression inside the bracket is an integer, say $$k$$. Thus $$a^n - b^n = (a-b)\,k$$, which means $$a - b$$ is a factor of $$a^n - b^n$$ for every positive integer $$n$$.

First find a general formula for $$(a+b)^6 - (a-b)^6$$. Expanding by the Binomial Theorem (coefficients $$1, 6, 15, 20, 15, 6, 1$$):

$$(a+b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6$$

$$(a-b)^6 = a^6 - 6a^5b + 15a^4b^2 - 20a^3b^3 + 15a^2b^4 - 6ab^5 + b^6$$

Subtracting, the even-power terms cancel and the odd-power terms double:

$$(a+b)^6 - (a-b)^6 = 2\left(6a^5b + 20a^3b^3 + 6ab^5\right)$$

Now put $$a = \sqrt{3}$$ and $$b = \sqrt{2}$$. Then

$$a^5 = (\sqrt{3})^5 = 9\sqrt{3},\quad a^3 = 3\sqrt{3},\quad a = \sqrt{3}$$

$$b^5 = (\sqrt{2})^5 = 4\sqrt{2},\quad b^3 = 2\sqrt{2},\quad b = \sqrt{2}$$

So the three products are:

$$a^5b = 9\sqrt{3}\cdot\sqrt{2} = 9\sqrt{6},\quad a^3b^3 = 3\sqrt{3}\cdot 2\sqrt{2} = 6\sqrt{6},\quad ab^5 = \sqrt{3}\cdot 4\sqrt{2} = 4\sqrt{6}$$

Substituting:

$$(\sqrt{3}+\sqrt{2})^6 - (\sqrt{3}-\sqrt{2})^6 = 2\left(6(9\sqrt{6}) + 20(6\sqrt{6}) + 6(4\sqrt{6})\right)$$

$$= 2\left(54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6}\right) = 2(198\sqrt{6}) = 396\sqrt{6}$$

First obtain a formula for $$(x+y)^4 + (x-y)^4$$. Expanding both fourth powers (coefficients $$1, 4, 6, 4, 1$$):

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

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

Adding, the odd-power terms cancel and the even-power terms double:

$$(x+y)^4 + (x-y)^4 = 2\left(x^4 + 6x^2y^2 + y^4\right)$$

Now put $$x = a^2$$ and $$y = \sqrt{a^2 - 1}$$. Then

$$x^4 = (a^2)^4 = a^8$$

$$y^2 = a^2 - 1,\qquad y^4 = (a^2-1)^2 = a^4 - 2a^2 + 1$$

$$x^2y^2 = (a^2)^2(a^2-1) = a^4(a^2-1) = a^6 - a^4$$

Substituting:

$$\left(a^2 + \sqrt{a^2-1}\right)^4 + \left(a^2 - \sqrt{a^2-1}\right)^4 = 2\left(a^8 + 6(a^6 - a^4) + (a^4 - 2a^2 + 1)\right)$$

$$= 2\left(a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1\right)$$

$$= 2\left(a^8 + 6a^6 - 5a^4 - 2a^2 + 1\right)$$

$$= 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2$$

Write $$0.99 = 1 - 0.01$$, so that $$(0.99)^5 = (1 - 0.01)^5$$.

By the Binomial Theorem the expansion of $$(1 - 0.01)^5$$ begins:

$$(1-0.01)^5 = \binom{5}{0}(1)^5 - \binom{5}{1}(1)^4(0.01) + \binom{5}{2}(1)^3(0.01)^2 - \cdots$$

Taking only the first three terms as an approximation:

$$(0.99)^5 \approx 1 - 5(0.01) + 10(0.01)^2$$

$$\approx 1 - 0.05 + 10(0.0001)$$

$$\approx 1 - 0.05 + 0.001$$

$$\approx 0.951$$

Hence $$(0.99)^5 \approx 0.951$$.

The Binomial Theorem expands a sum of two quantities, so we group the three terms into two. Take $$A = 1 + \dfrac{x}{2}$$ and $$B = \dfrac{2}{x}$$, so the expression is $$(A - B)^4$$.

Expanding $$(A-B)^4$$ with coefficients $$1, 4, 6, 4, 1$$:

$$(A-B)^4 = A^4 - 4A^3B + 6A^2B^2 - 4AB^3 + B^4$$

Since $$B = \dfrac{2}{x}$$, we have $$B^2 = \dfrac{4}{x^2}$$, $$B^3 = \dfrac{8}{x^3}$$, $$B^4 = \dfrac{16}{x^4}$$, so

$$(A-B)^4 = A^4 - \dfrac{8A^3}{x} + \dfrac{24A^2}{x^2} - \dfrac{32A}{x^3} + \dfrac{16}{x^4}$$

Now expand the powers of $$A = 1 + \dfrac{x}{2}$$ by the Binomial Theorem:

$$A^2 = 1 + x + \dfrac{x^2}{4}$$

$$A^3 = 1 + \dfrac{3x}{2} + \dfrac{3x^2}{4} + \dfrac{x^3}{8}$$

$$A^4 = 1 + 2x + \dfrac{3x^2}{2} + \dfrac{x^3}{2} + \dfrac{x^4}{16}$$

Substitute these into the five terms one at a time:

$$A^4 = 1 + 2x + \dfrac{3x^2}{2} + \dfrac{x^3}{2} + \dfrac{x^4}{16}$$

$$-\dfrac{8A^3}{x} = -\dfrac{8}{x}\left(1 + \dfrac{3x}{2} + \dfrac{3x^2}{4} + \dfrac{x^3}{8}\right) = -\dfrac{8}{x} - 12 - 6x - x^2$$

$$\dfrac{24A^2}{x^2} = \dfrac{24}{x^2}\left(1 + x + \dfrac{x^2}{4}\right) = \dfrac{24}{x^2} + \dfrac{24}{x} + 6$$

$$-\dfrac{32A}{x^3} = -\dfrac{32}{x^3}\left(1 + \dfrac{x}{2}\right) = -\dfrac{32}{x^3} - \dfrac{16}{x^2}$$

$$B^4 = \dfrac{16}{x^4}$$

Finally collect like terms (constant: $$1 - 12 + 6 = -5$$; coefficient of $$x$$: $$2 - 6 = -4$$; coefficient of $$x^2$$: $$\tfrac{3}{2} - 1 = \tfrac{1}{2}$$; coefficient of $$\tfrac{1}{x}$$: $$-8 + 24 = 16$$; coefficient of $$\tfrac{1}{x^2}$$: $$24 - 16 = 8$$):

$$\left(1 + \dfrac{x}{2} - \dfrac{2}{x}\right)^4 = \dfrac{x^4}{16} + \dfrac{x^3}{2} + \dfrac{x^2}{2} - 4x - 5 + \dfrac{16}{x} + \dfrac{8}{x^2} - \dfrac{32}{x^3} + \dfrac{16}{x^4}$$

To use the Binomial Theorem we group the three terms into two. Write $$p = 3x^2 - 2ax$$, so the expression becomes $$(p + 3a^2)^3$$.

Expanding $$(p + 3a^2)^3$$ with coefficients $$1, 3, 3, 1$$:

$$(p + 3a^2)^3 = p^3 + 3p^2(3a^2) + 3p(3a^2)^2 + (3a^2)^3$$

$$= p^3 + 9a^2 p^2 + 27a^4 p + 27a^6$$

Now compute the powers of $$p = 3x^2 - 2ax$$ by the Binomial Theorem.

$$p^2 = (3x^2 - 2ax)^2 = 9x^4 - 12ax^3 + 4a^2x^2$$

$$p^3 = (3x^2 - 2ax)^3 = (3x^2)^3 - 3(3x^2)^2(2ax) + 3(3x^2)(2ax)^2 - (2ax)^3$$

$$p^3 = 27x^6 - 54ax^5 + 36a^2x^4 - 8a^3x^3$$

Substitute into the four pieces:

$$p^3 = 27x^6 - 54ax^5 + 36a^2x^4 - 8a^3x^3$$

$$9a^2 p^2 = 9a^2(9x^4 - 12ax^3 + 4a^2x^2) = 81a^2x^4 - 108a^3x^3 + 36a^4x^2$$

$$27a^4 p = 27a^4(3x^2 - 2ax) = 81a^4x^2 - 54a^5x$$

$$(3a^2)^3 = 27a^6$$

Add the four pieces and collect like terms:

$$x^6:\ 27x^6 \qquad x^5:\ -54ax^5$$

$$x^4:\ 36a^2x^4 + 81a^2x^4 = 117a^2x^4$$

$$x^3:\ -8a^3x^3 - 108a^3x^3 = -116a^3x^3$$

$$x^2:\ 36a^4x^2 + 81a^4x^2 = 117a^4x^2$$

$$x:\ -54a^5x \qquad \text{constant}:\ 27a^6$$

Therefore

$$(3x^2 - 2ax + 3a^2)^3 = 27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6$$