Let $f(x) =x^5+x^2+1$ with $a, b, c, d, e$ as zeros. . . Let $f(x) =x^5+x^2+1$ with $a, b, c, d, e$ as zeros and $g(x)=x^2-2.$ Show that $$g(a)g(b)g(c)g(d)g(e)=-23. $$
I have no idea how to solve it. 
Please help. 
Thanks in advance. 

Let $a_1,\ldots,a_5$ be the roots of $f(x)=x^5+x^2+1$. Then for every real $x$ we have:
$$f(x)=x^5+x^2+1=\prod_{i=1}^5(x-a_i).$$
So, $g(a_1)\cdot g(a_2)\cdot g(a_3)\cdot g(a_4)\cdot g(a_5)$ is equal to:
\begin{align*}
\prod_{i=1}^5 g(a_i)&=\prod_{i=1}^5 (a_i^2-2)=\prod_{i=1}^5 (a_i-\sqrt{2})(a_i+\sqrt{2})\\
&=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot \prod_{i=1}^5 (a_i-\sqrt{2})\\
&=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot \prod_{i=1}^5 [(-1)(\sqrt{2}-a_i)]\\
&=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot (-1)^5 \prod_{i=1}^5 (\sqrt{2}-a_i)\\
&=\prod_{i=1}^5 [(-1)(a_i+\sqrt{2})] \cdot \prod_{i=1}^5 (\sqrt{2}-a_i)\\
&=\prod_{i=1}^5 (-\sqrt{2}-a_i) \cdot \prod_{i=1}^5 (\sqrt{2}-a_i)\\
&=f(-\sqrt{2})\cdot f(\sqrt{2})\\
&=(-4\sqrt{2}+3)(4\sqrt{2}+3)\\
&=-23
\end{align*} 
