## Precalculus (6th Edition) Blitzer

The simplest form of the expression is ${{\left( {{x}^{2}}-1 \right)}^{4}}={{x}^{8}}-4{{x}^{6}}+6{{x}^{4}}-4{{x}^{2}}+1$
Thus, we get \begin{align} & {{\left( {{x}^{2}}-1 \right)}^{4}}=\left( \begin{matrix} 4 \\ 0 \\ \end{matrix} \right){{\left( {{x}^{2}} \right)}^{4}}+\left( \begin{matrix} 4 \\ 1 \\ \end{matrix} \right){{\left( {{x}^{2}} \right)}^{3}}\left( -1 \right)+\left( \begin{matrix} 4 \\ 2 \\ \end{matrix} \right){{\left( {{x}^{2}} \right)}^{2}}{{\left( -1 \right)}^{2}}+\left( \begin{matrix} 4 \\ 3 \\ \end{matrix} \right)\left( {{x}^{2}} \right){{\left( -1 \right)}^{3}}+\left( \begin{matrix} 4 \\ 4 \\ \end{matrix} \right){{\left( -1 \right)}^{4}} \\ & =\frac{4!}{0!\left( 4-0 \right)!}\left( {{x}^{8}} \right)+\frac{4!}{1!\left( 4-1 \right)!}\left( -{{x}^{6}} \right)+\frac{4!}{2!\left( 4-2 \right)!}\left( {{x}^{4}} \right)+\frac{4!}{3!\left( 4-3 \right)!}\left( -{{x}^{2}} \right)+\frac{4!}{4!\left( 4-4 \right)!}\left( 1 \right) \\ & ={{x}^{8}}-4{{x}^{6}}+6{{x}^{4}}-4{{x}^{2}}+1 \end{align}