## Precalculus (6th Edition) Blitzer

a) 25 b) -5 c) $-x^3+x+1$ d) $27{{a}^{3}}-3a+1$
a) Putting the value of $x=3$ in the equation as follows: \begin{align} & h\left( 3 \right)={{3}^{3}}-3+1 \\ & =27-3+1 \\ & =25 \end{align} b) Putting the value of $x=-2$ in the equation as follows: \begin{align} & h\left( -2 \right)={{\left( -2 \right)}^{3}}-\left( -2 \right)+1 \\ & =-8+2+1 \\ & =-5 \end{align} c) Putting the value of $x=-x$ in the equation as follows: \begin{align} & h\left( -x \right)={{\left( -x \right)}^{3}}-\left( -x \right)+1 \\ & =-{{x}^{3}}+x+1 \end{align} d) Putting the value of $x=3a$ in the equation as follows: \begin{align} & h\left( 3a \right)={{\left( 3a \right)}^{3}}-\left( 3a \right)+1 \\ & =27{{a}^{3}}-3a+1 \end{align}