Roots of biquadratic equation 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.
Complete solution of the problem:

Given that α,β,γ &amp; δ are squares of the roots of $x^4+bx^2+cx+d=0$ hence $\sqrt{α},\sqrt{β},\sqrt{γ}$  &amp; $\sqrt{δ}$ are the roots of the given equation. 
Thus, we have

sum of the roots
$$=\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ}=\frac{-B}{A}=\frac{0}{1}=0$$  &amp;

sum of the products of two roots
$$= \sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ}=\frac{C}{A}=\frac{b}{1}=b$$
now, taking the square of the first expression, we get
 $$(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})^2=0 $$
 $$α+β+γ+δ+2(\sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ})=0$$

$$∑α+2b=0 $$  $∑α=-2b\tag1$       

Now, the product of the roots
$$=\sqrt{α}  \sqrt{β} \sqrt{γ} \sqrt{δ}=\frac{E}{A}=\frac{d}{1}=d$$
$$ \sqrt{αβγδ}=d   $$ $$  αβγδ=d^2  \tag2$$

Now, satisfying the given equation $x^4+bx^2+cx+d=0$ by its roots $\sqrt{α},\sqrt{β},\sqrt{γ}$  &amp; $\sqrt{δ}$  one by one, we get the following
$$\begin{align}
α^2+bα+c\sqrt{α}+d=0\\
β^2+bβ+c\sqrt{β}+d=0\\
γ^2+bγ+c\sqrt{γ}+d=0\\
δ^2+bδ+c\sqrt{δ}+d=0\\
\end{align}$$
Now by adding above four expressions column-wise, we get
$$α^2+β^2+γ^2+δ^2+b(α+β+γ+δ)+c(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})+4d=0$$

$$α^2+β^2+γ^2+δ^2+b(-2b)+c(0)+4d=0$$
  $$α^2+β^2+γ^2+δ^2=2b^2-4d \tag3$$

Now by squaring both the sides of eq(1),we have
 $$(α+β+γ+δ)^2=(-2b)^2$$
$$α^2+β^2+γ^2+δ^2+2(αβ+αγ+αδ+βγ+βδ+γδ)=4b^2$$
By substituting the value from eq(3), we get
$$2b^2-4d+2∑αβ=4b^2 $$
$$2∑αβ= 2b^2+4d $$
$$  4∑αβ= 4b^2+8d \tag4$$ 

Now, by substituting all the corresponding values from eq(1), (2) &amp; (4) in the LHS of the expression to be proved, we get

$$LHS=64αβγδ-[4∑αβ-(∑α)^2 ]^2=64d^2-[4b^2+8d-(-2b)^2 ]^2$$
$$ =64d^2-[8d]^2=64d^2-64d^2=0=RHS  $$