Question about Vieta's formula Suppose $\alpha,\beta,\gamma$ are the roots of equation $x^3-4x+2=0,$ how to find $[(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)]^2?$
Indeed, given an expression , what is the trick to change the given expression into an expression of Vieta's formula?
 A: You just have to expand $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$ to get the expression in terms of sum of products of roots and you can use Vieta's formula. Square it to get the quantity that you want.
Also, for an equation $$y^3+py+q=0\tag{1}$$
It is known that 
$$[(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)]^2=-27q^2-4p^3$$
This is known as the discriminant function to $(1)$. Try to verify it.
Edit:
\begin{align}
&[(\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2)-(\alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha)]^2\\
&=[(\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2)+(\alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha)]^2 - 4(\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2)(\alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha)
\end{align}
Focusing on the first term inside the square:
\begin{align}
&(\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2)+(\alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha)\\
&=\alpha\beta(\alpha+\beta)+\beta\gamma(\beta+\gamma)+\alpha\gamma(\alpha+\gamma)\\&= \alpha\beta(\alpha+\beta+\gamma)+\beta\gamma(\beta+\gamma+\alpha)+\alpha\gamma(\alpha+\gamma+\beta)-3\alpha\beta\gamma\\
&=(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\alpha\gamma)-3\alpha\beta\gamma
\end{align}
Now, let's move on to the second term ignoring the $4$,
\begin{align}
&(\alpha\beta^2+\beta\gamma^2+\gamma\alpha^2)(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha)
\\&=\alpha^3\beta^3+\alpha\beta^4\gamma+\alpha^2\beta^2\gamma^2\\&+\alpha^2\beta^2\gamma^2+\beta^3\alpha^3+\alpha\beta\gamma^4\\
&+\alpha^4\beta\gamma + \alpha^2\beta^2\gamma^2+\alpha^3\beta^3 \\
&= \alpha^3\beta^3+\beta^3\gamma^3+\gamma^3\alpha^3+3(\alpha\beta\gamma)^2+\alpha\beta\gamma(\alpha^3+\beta^3+\gamma^3)
\end{align}
To finish it up, try to explore the formula $(a+b+c)^3.$
A: Hint: Use that $$\alpha+\beta+\gamma=0$$
$$\alpha\beta+\alpha\gamma+\beta\gamma=-4$$
$$\alpha\beta\gamma=-2$$
