$\displaystyle x^{3}+mx+n=0$ 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

$\displaystyle det=(x_1-x_2)(x_1-x_3)(x_2-x_3)$

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).

$\displaystyle det=(x_3-x_2)(3x_3x_2+2m)$

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 pre-algebra too. This is a linear algebra problem, but it's for highschool.