Binomial Expansion of $$(a+b)^n+(a-b)^n$$

To find the degree of the polynomial, we expand the expression completely using the Binomial Theorem. This question contains a deliberate trap: the highest power ($$x^{12}$$) completely cancels out during simplification.

Step 1: Apply the Binomial Theorem expansion
Let $$a = x^2 - 3$$ and $$b = (1 - x^4)^{\frac{1}{2}}$$. The expression is in the form $$(a + b)^6 + (a - b)^6$$.

Expanding both parts with the Binomial Theorem and adding them together cancels out all terms with odd powers of $$b$$:
$$(a+b)^6 + (a-b)^6 = 2 \left[ \binom{6}{0}a^6 + \binom{6}{2}a^4b^2 + \binom{6}{4}a^2b^4 + \binom{6}{6}b^6 \right]$$

Substitute the combination values $$\binom{6}{0}=1$$, $$\binom{6}{2}=15$$, $$\binom{6}{4}=15$$, and $$\binom{6}{6}=1$$:
$$= 2 [ a^6 + 15a^4b^2 + 15a^2b^4 + b^6 ]$$

Step 2: Write out the full expansion of each component
Substitute $$a = x^2 - 3$$ and $$b^2 = 1 - x^4$$ back into the formula:
$$= 2 [ (x^2 - 3)^6 + 15(x^2 - 3)^4(1 - x^4) + 15(x^2 - 3)^2(1 - x^4)^2 + (1 - x^4)^3 ]$$

We fully expand each of the four components inside the brackets:

1. For $$(x^2 - 3)^6$$:
$$x^{12} - 18x^{10} + 135x^8 - 540x^6 + 1215x^4 - 1458x^2 + 729$$

2. For $$15(x^2 - 3)^4(1 - x^4)$$:
$$15(x^8 - 12x^6 + 54x^4 - 108x^2 + 81)(1 - x^4)$$
$$= -15x^{12} + 15x^8 + 180x^{10} - 180x^6 - 810x^8 + 810x^4 + 1620x^6 - 1620x^2 - 1215x^4 + 1215$$
$$= -15x^{12} + 180x^{10} - 795x^8 + 1440x^6 - 405x^4 - 1620x^2 + 1215$$

3. For $$15(x^2 - 3)^2(1 - x^4)^2$$:
$$15(x^4 - 6x^2 + 9)(x^8 - 2x^4 + 1)$$
$$= 15x^{12} - 30x^8 + 15x^4 - 90x^{10} + 180x^6 - 90x^2 + 135x^8 - 270x^4 + 135$$
$$= 15x^{12} - 90x^{10} + 105x^8 + 180x^6 - 255x^4 - 90x^2 + 135$$

4. For $$(1 - x^4)^3$$:
$$-x^{12} + 3x^8 - 3x^4 + 1$$

Step 3: Combine all terms to get the complete polynomial expression
Now, sum all four fully expanded parts together inside the brackets:
$$\text{Sum} = (x^{12} - 15x^{12} + 15x^{12} - x^{12}) + (-18x^{10} + 180x^{10} - 90x^{10}) + (135x^8 - 795x^8 + 105x^8 + 3x^8) + (-540x^6 + 1440x^6 + 180x^6) + (1215x^4 - 405x^4 - 255x^4 - 3x^4) + (-1458x^2 - 1620x^2 - 90x^2) + (729 + 1215 + 135 + 1)$$

Simplify each grouped power step-by-step:
* For $$x^{12}$$: $$1 - 15 + 15 - 1 = 0$$
* For $$x^{10}$$: $$-18 + 180 - 90 = 72$$
* For $$x^8$$: $$135 - 795 + 105 + 3 = -552$$
* For $$x^6$$: $$-540 + 1440 + 180 = 1080$$
* For $$x^4$$: $$1215 - 405 - 255 - 3 = 552$$
* For $$x^2$$: $$-1458 - 1620 - 90 = -3168$$
* For constants: $$729 + 1215 + 135 + 1 = 2080$$

Multiplying the entire simplified bracket by the front factor of $$2$$ gives the complete expanded polynomial:
$$\text{Complete Expression} = 2(72x^{10} - 552x^8 + 1080x^6 + 552x^4 - 3168x^2 + 2080)$$
$$\text{Complete Expression} = 144x^{10} - 1104x^8 + 2160x^6 + 1104x^4 - 6336x^2 + 4160$$

Since the highest remaining power of $$x$$ is 10, the true degree of the polynomial is 10.

Observe that

$$3^{10-r}=(\sqrt3)^{20-2r}$$

Therefore

$$S=\sum_{r=0}^{10}\binom{20}{2r}(\sqrt3)^{20-2r}(-1)^r$$

Now use the standard identity

$$\frac{(a+b)^{20}+(a-b)^{20}}{2}=\sum_{r=0}^{10}\binom{20}{2r}a^{20-2r}b^{2r}$$

Take

$$a=\sqrt3,\qquad b=i$$

Since

$$i^{2r}=(-1)^r$$

we get

$$S=\frac{(\sqrt3+i)^{20}+(\sqrt3-i)^{20}}{2}$$

Now write

$$\sqrt3+i=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)$$

Hence by De Moivre's theorem,

$$\left(\sqrt3+i\right)^{20}=2^{20}\left(\cos\frac{20\pi}{6}+i\sin\frac{20\pi}{6}\right)$$

Similarly,

$$\left(\sqrt3-i\right)^{20}=2^{20}\left(\cos\frac{20\pi}{6}-i\sin\frac{20\pi}{6}\right)$$

Adding,

$$S=\frac{2^{20}\left(2\cos\frac{20\pi}{6}\right)}{2}$$

$$=2^{20}\cos\frac{20\pi}{6}$$

$$=2^{20}\cos\frac{10\pi}{3}$$

$$=2^{20}\cos\frac{4\pi}{3}$$

$$=2^{20}\left(-\frac12\right)$$

$$=-2^{19}$$

Final Answer :

$$-2^{19}$$