Answer
$1 -6x^{2}+15x^{4}-20x^{6}+15x^{8}-6x^{10}+x^{12}$
Work Step by Step
The Binomial Theorem$:$
$(a+b)^{n}=\left(\begin{array}{l}
n\\
0
\end{array}\right)a^{n}+\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}b+\left(\begin{array}{l}
n\\
2
\end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l}
n\\
n
\end{array}\right)b^{n}$
----------------
$\left(\begin{array}{l}
6\\
0
\end{array}\right)1^{6}\cdot(-x^{2})^{0}=1\cdot 1\cdot 1=1$
$\left(\begin{array}{l}
6\\
1
\end{array}\right)1^{5}\cdot(-x^{2})^{1}=6\cdot 1\cdot(-x^{2})=-6x^{2}$
$\displaystyle \left(\begin{array}{l}
6\\
2
\end{array}\right)1^{4}\cdot(-x^{2})^{2}=\frac{6(5)}{1(2)}\cdot x^{4}=15x^{4}$
$\displaystyle \left(\begin{array}{l}
6\\
3
\end{array}\right)1^{3}\cdot(-x^{2})^{3}=\frac{6(5)(4)}{1(2)(3)}\cdot(-x^{2})^{3}=-20x^{6}$
$\displaystyle \left(\begin{array}{l}
6\\
4
\end{array}\right)1^{2}\cdot(-x^{2})^{4}=\frac{6(5)(4)(3)}{1(2)(3)(4)}\cdot(-x^{2})^{4}=15x^{8}$
$\left(\begin{array}{l}
6\\
5
\end{array}\right)1^{1}\cdot(-x^{2})^{5}=6(-1)\cdot x^{10}=-6x^{10}$
$\left(\begin{array}{l}
6\\
6
\end{array}\right)1^{0}\cdot(-x^{2})^{6}=1\cdot x^{12}=x^{12}$
$(1-x^{2})=1-6x^{2}+15x^{4}-20x^{6}+15x^{8}-6x^{10}+x^{12}$
