<s>[INST] $x_1,x_2,x_3$ the roots of $x^3+mx+n=0$ Find determinant of $A^{2}$ Let $m,n\in\mathbb{R}; x_1,x_2,x_3 $ the roots of $x^3+mx+n=0$ and  the matrix $A=\begin{pmatrix}
1 & 1 &1 \\ 
x_1 &x_2  &x_3 \\ 
x^{2}_1 & x^{2}_2 & x^{2}_3
\end{pmatrix}$
I need to find determinant of $A^2$ which is $det(A)\cdot det(A)$
I got $det(A)=(x_2-x_1)(x_3-x_1)(x_3-x_2)$.I know that $x_1+x_2+x_3=0$, $x_1x_2+x_1x_3+x_2x_3=m$, $x_1x_2x_3=-n$.
I expanded the determinant and I tried to factorize but I can't use Vieta formula because I don't get a sum or a product.
Also I tried to find $det(A\cdot A^T)$ but the calculations are very heavy.
How to approach the exercise?
 [/INST] An alternative method!
You found:
$$\det(A)=(x_2-x_1)(x_3-x_1)(x_3-x_2), \\
x_1+x_2+x_3=0, x_1x_2+x_1x_3+x_2x_3=m, x_1x_2x_3=-n$$
Note: 
$$x_1+x_2+x_3=0 \Rightarrow x_1+x_2=-x_3 \Rightarrow x_1^2+x_2^2=x_3^2-2x_1x_2 \ \ (1)\\
$$
Also note the famous formula:
$$x^3+y^3+z^3=3xyz+(x+y+z)[(x+y+z)^2-3(xy+yz+zx)] \ (2) \ \ \text{or}\\
(xy)^3+(yz)^3+(zx)^3=3(xyz)^2+(xy+yz+zx)[(xy+yz+zx)^2-3xyz(x+y+z)] $$
You need to find:
$$\begin{align}\det(A)^2&=(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2=\\
&=(x_1^2+x_2^2-2x_1x_2)(x_1^2+x_3^2-2x_1x_3)(x_2^2+x_3^2-2x_2x_3)\stackrel{(1)}=\\
&=(x_3^2-4x_1x_2)(x_2^2-4x_1x_3)(x_1^2-4x_2x_3)=\\
&=-63x_1^2x_2^2x_3^2-4[(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3]+16x_1x_2x_3(x_1^3+x_2^3+x_3^3)\stackrel{(2)}=\\
&=-63n^2-4[3n^2+m(m^2-0)]-16n(0-3n)=\\
&=-27n^2-4m^3.\end{align}$$
</s>