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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?

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    $\begingroup$ $det(A)^2$ is a symmetric polynomial (i.e. remains the same if you interchange $x_i$ and $x_j$) . So, you should have that $det(A)^2$ can be expressed as a polynomial of the elementary symmetric polynomials $x_1+x_2+x_3$, $x_1x_2+x_2x_3+x_3x_1$ and $x_1x_2x_3$. $\endgroup$ – Julian Mejia May 25 at 11:21
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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: $$\begin{align}\det(A)^2&=(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2=\\ &=(x_1^2+x_2^2-2x_1x_2)(x_1^2+x_3^2-2x_1x_3)(x_2^2+x_3^2-2x_2x_3)\stackrel{(1)}=\\ &=(x_3^2-4x_1x_2)(x_2^2-4x_1x_3)(x_1^2-4x_2x_3)=\\ &=-63x_1^2x_2^2x_3^2-4[(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3]+16x_1x_2x_3(x_1^3+x_2^3+x_3^3)\stackrel{(2)}=\\ &=-63n^2-4[3n^2+m(m^2-0)]-16n(0-3n)=\\ &=-27n^2-4m^3.\end{align}$$

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  • $\begingroup$ Thank you very much for your response! $\endgroup$ – DaniVaja May 25 at 13:32
  • $\begingroup$ my pleasure, good luck! $\endgroup$ – farruhota May 25 at 14:55
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Hint: $\det(A^2) = \det(A)^2 = (x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$ is the discriminant of $x^3+mx+n$.

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    $\begingroup$ I think it is "too obvious" for that Vandermonde and discriminant will settle this. Therefore I would have searched the site rather thoroughly before posting anything that answers a low hanging fruit. Particularly as a relative veteran of the site. $\endgroup$ – Jyrki Lahtonen May 25 at 11:28
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You know that$$\det(A^2)=\det\nolimits^2A=\bigl((x_1-x_2)(x_2-x_3)(x_1-x_3)\bigr)^2,$$which is a symmetric polynomial. So, you can write it as a polynomial in the elementary symmetric polynomials. It turns out that it is equal to:\begin{multline}-27{x_1}^2{x_2}^2{x_3}^2-4{x_1}{x_2}{x_3}({x_1}+{x_2}+{x_3})^3+({x_1}{x_2}+{x_1}{x_3}+{x_2}{x_3})^2 ({x_1}+{x_2}+{x_3})^2+\\+18{x_1}{x_2}{x_3}({x_1}{x_2}+{x_1}{x_3}+{x_2}{x_3}) ({x_1}+{x_2}+{x_3})-4 ({x_1}{x_2}+{x_1}{x_3}+{x_2}{x_3})^3.\end{multline}But you know that $x_1+x_2+x_3=0$, that $x_1x_2+x_2x_3+x_1x_3=m$, and that $x_1x_2x_3=-n$. So…

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  • $\begingroup$ I think I don't understand.The first part I get it.Then I expand in paratheses and I square?I don't understand which formula you applied to get $-27x^{2}_1...$ $\endgroup$ – DaniVaja May 25 at 11:39
  • $\begingroup$ I used the algorithme sketched here in order to express $\bigl((x_1-x_2)(x_2-x_3)(x_1-x_3)\bigr)^2$ as a polynomial in $x_1x_2x_3$, $x_1x_2+x_2x_3+x_1x_3$, and $x_1+x_2+x_3$. $\endgroup$ – José Carlos Santos May 25 at 11:49

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