<s>[INST] Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$ Let $\alpha$, $\beta$, $\gamma$, $\delta$ be the roots of $$z^4-2z^3+z^2+z-7=0$$ then find value of $$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$$
Are Vieta's formulas appropriate?
 [/INST] Rearrange the equation to $z(2z^2-1)=z^4+z^2-7$ square this & substitute $y=z^2+1$ 
\begin{eqnarray*}
(y-1)(2y-3)^2=((y-1)^2+(y-1)-7)^2 \\
y^4+\cdots +58 =0.
\end{eqnarray*}
Note that the roots of this polynomial will be $\alpha^2+1,\beta^2+1,\gamma^2+1,\delta^2+1$ and we have not calculated all the terms explicitly (only the ones required to obtain the product of these roots.)
So $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=\color{blue}{58}$.
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