
Determinant
$\displaystyle x^{3}+mx+n=0$ is given with m and n real numbers and $\displaystyle x_1, x_2, x_3$ it's solutions.
It's asked for the value of the determinant:
$\displaystyle det = \begin{vmatrix}
1 & 1 & 1\\
x_1 & x_2 & x_3 \\
x_1^{2} & x_2^{2} & x_3^{2}
\end{vmatrix}$
I have tried to solve it in the Vandermonde way
$\displaystyle det=(x_1x_2)(x_1x_3)(x_2x_3)$
Or to sum up columns 2 and 3 to the first (to obtain the sum, and the sum of squares of the solutions  which are found easily with Viete).
$\displaystyle det=(x_3x_2)(3x_3x_2+2m)$
I have managed to find out that:
$\displaystyle (x_1^{2}+x_1x_2+x_2^{2})(x_1^{2}+x_1x_3+x_3^{2})(x _2^{2}+x_2x_3+x_3^{2})=m$
the value of the determinant beeing $\displaystyle \pm \sqrt{4m^{3}27n^{2}}$ thinking that I should sqare the first identety and make some conection with this one.
I'm sorry for posting this problem to prealgebra too. This is a linear algebra problem, but it's for highschool.

Let the roots of $\displaystyle x^3+mx+n=0$ be $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$. Then $\displaystyle a+b+c=0$, $\displaystyle bc+ca+ab=m$ and $\displaystyle abc=n$.
Here's one way to find $\displaystyle (ab)(bc)(ca)$, probably not the best.
Start with $\displaystyle a^2+b^2+c^2=(a+b+c)^22(bc+ca+ab)=2m$, and so $\displaystyle a^2+b^2=2mc^2$.
Since $\displaystyle c$ is a root, $\displaystyle c^3=mcn=mc+abc$, so $\displaystyle c^2=abm$. So now $\displaystyle a^2+b^2=mab$.
Therefore $\displaystyle (ab)^2=a^2+b^22ab=m3ab$. Similarly $\displaystyle (bc)^2=m3bc$ and $\displaystyle (ca)^2=m3ca$.
Thus $\displaystyle (ab)^2(bc)^2(ca)^2=(m+3ab)(m+3bc)(m+3ca)$ $\displaystyle =(27a^2b^2c^2+9m(a^2bc+b^2ca+c^2ab)+3m^2(ab+bc+ca)+ m^3)$.
Using the root relations, this last quantity is $\displaystyle (27n^2+4m^3)$. Hope this helps.

It was very helpful. I have found out, in other place, antoher method.
Thank you very much, this one I have understood it.

I will post the other solution, it's not so complicated.
Let it be
$\displaystyle A= \begin{pmatrix}
1 & 1 & 1\\
x_1 & x_2 & x_3\\
x_1^{2} & x_2^{2} & x_3^{2}
\end{pmatrix}, A^{t}=\begin{pmatrix}
1 & x_1 & x_1^{2}\\
1 & x_2 & x_2^{2}\\
1 & x_3 & x_3^{2}
\end{pmatrix}, A^{tr}transposed \, matrix$
hope transposed is the name in English.
$\displaystyle det(AA^{tr})=det(A)det(A^{tr})=det(A)^{2}=\begin{v matrix}
1 & x_1+x_2+x_3 & x_1^{2}+x_2^{2}+x_3^{2}\\
x_1+x_2+x_3 &x_1^{2}+x_2^{2}+x_3^{2} &x_1^{3}+x_2^{3}+x_3^{3} \\
x_1^{2}+x_2^{2}+x_3^{2} & x_1^{3}+x_2^{3}+x_3^{3} & x_1^{4}+x_2^{4}+x_3^{4}
\end{vmatrix}$
From Viete's relations:
$\displaystyle x_1+x_2+x_3=0 $
$\displaystyle x_1^{2}+x_2^{2}+x_3^{2}=(x_1+x_2+x_3)^22(x_1x_2+x_1x_3+x_2x_3)=02m $
$\displaystyle x^{3}=mxn $
$\displaystyle x^{4}=mx^2nx $
$\displaystyle x_1^{3}+x_2^{3}+x_3^{3}=m(x_1+x_2+x_3)3n $
$\displaystyle x_1^{4}+x_2^{4}+x_3^{4}=m(x_1^{2}+x_2^{2}+x_3^{2})n(x_1+x_2+x_3) $
I didn't solve it. Someone from another forum (from my country) did. Hope it helps others too.