This question also was a part of my today's maths olympiad paper:
If squares of the roots of $x^4 + bx^2 + cx + d = 0$ are $\alpha, \beta, \gamma, \delta$ then prove that: $64\alpha\beta\gamma\delta - [4\Sigma \alpha\beta - (\Sigma \alpha)^2]^2 = 0$
I found value of $\alpha\beta\gamma\delta$ = $d^2$ and $\Sigma \alpha$ = -2b. How to move on ?
Set $y=x^2$ and find an equation for $y$.
To start with $y^2+by+cx+d=0$ so that $c^2y=c^2x^2=(y^2+by+d)^2=y^4+2by^3+(2d+b^2)y^2+2dy+d^2$ Whence $$y^4+2by^3+(2d+b^2)y^2+(2d-c^2)y+d^2=0$$
The equation for $y$ has roots which are the squares of the roots of the original equation. You can use Vieta's relations with this.