Answer
$x^{12}-6x^{10}+15x^8-20x^6+15x^4-6x^2+1$
Work Step by Step
We are given the expression $(1-x^2)^6$.
For expanding it we will use the $\textit{Binomial Theorem}$:
$$\begin{align}(a+b)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+\cdots+\binom{n}{n}b^n.\end{align}\tag1$$
Substitute $a=1$, $b=-x^2$, $n=6$ in Eq. $(1)$:
$$\begin{align*}
(1-x^2)^6&=\binom{6}{0}1^6+\binom{6}{1}1^5(-x^2)+\binom{6}{2}1^4(-x^2)^2\\
&\phantom{=}+\binom{6}{3}1^3(-x^2)^3+\binom{6}{4}1^2(-x^2)^4\\
&\phantom{=}+\binom{6}{5}1^1(-x^2)^5+\binom{6}{6}(-x^2)^6\\
&=1-6x^2+15x^4-20x^6+15x^8-6x^{10}+x^{12}
\end{align*}$$
The expansion is:
$$(1-x^2)^6=x^{12}-6x^{10}+15x^8-20x^6+15x^4-6x^2+1.$$
