Solve the following question

Question

If $\alpha$ ,$\beta$, $\gamma$ and $\delta$ be the roots of the equation $x^4+px^3+qx^2+rx+s=0$, show that $(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)$=$(1-q+s)^2 +(p-r)^2$. I tried to solve this by substituting values from the given equation to the RHS, but the calculations are just too long.I ,however, know that there is some other way to do it but I simply don't know what it is.Please help me.

Answer 1

HInt: Use these facts:

• $s=\alpha\beta\gamma\delta$;
• $r=-\beta\gamma\delta-\alpha\gamma\delta-\alpha\beta\delta-\alpha\beta\gamma$;
• $q=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta$;
• $p=-\alpha-\beta-\gamma-\delta$.

Answer 2

Hint. We have that $$f(x)=x^4+px^3+qx^2+rx+s=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta).$$ Then, since $(i-a)(-i-a)=-i^2+a^2=1+a^2$, it follows that $$(1+\alpha^2)(1+\beta ^2)(1+\gamma^2)(1+\delta^2)=f(i)f(-i).$$