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Factorize and prove that

$$ \Delta=\begin{vmatrix} yz-x^2&zx-y^2&xy-z^2\\ zx-y^2&xy-z^2&yz-x^2\\ xy-z^2&yz-x^2&zx-y^2 \end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2 $$ using factor theorem.

My Attempt:

$\Delta$ is a homogeneous symmetric polynomial of degree $6$.

When $(x-y)^2+(y-z)^2+(z-x)^2=0$, i.e. $x=y=z$ $$ \Delta=\begin{vmatrix} 0&0&0\\ 0&0&0\\ 0&0&0\\ \end{vmatrix}=0 $$ Thus, $(x-y)^2+(y-z)^2+(z-x)^2$ is a factor.

How do I extract the other $(x-y)^2+(y-z)^2+(z-x)^2$ from $\Delta$ $\color{red}{?}$

Does this have anything to do with all rows (or columns) being zero when $(x-y)^2+(y-z)^2+(z-x)^2=0$ $\color{red}{?}$

If I can extract that then i think I know how to proceed. The remaining factor must be a homogeneous quadratic symmetric polynomial, i.e. $p(x,y,z)=a(x^2+y^2+z^2)+b(xy+yz+zx)$ $$ \Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.a(x^2+y^2+z^2)+b(xy+yz+zx) $$ $$ \Delta(1,0,0)=\begin{vmatrix} -1&0&0\\ 0&0&-1\\ 0&-1&0\\ \end{vmatrix}=1=4.a\implies a=\frac{1}{4} $$ $$ \Delta(1,1,0)=\begin{vmatrix} -1&-1&1\\ -1&1&-1\\ 1&-1&-1\\ \end{vmatrix}=\begin{vmatrix} 0&0&1\\ -2&0&-1\\ 0&-2&-1\\ \end{vmatrix}\\ =\begin{vmatrix} -2&0\\ 0&-2\\ \end{vmatrix}=4=4.(2a+b)=4(1/2+b)=2+4b\\ \implies b=\frac{1}{2} $$ $$ \Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.\frac{1}{4}(x^2+y^2+z^2)+\frac{1}{2}(xy+yz+zx)\\ =\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.(x^2+y^2+z^2+2xy+2yz+2zx)\\ =\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2(x+y+z)^2 $$ Note:

I am trying to factorize the determinant using factor theorem given the fact that the determinant is a homogeneous symmetric polynomial of degree 6.

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    $\begingroup$ How do you infer that $(x-y)^2+(y-z)^2+(z-x)^2$ is a factor? Couldn't you use the same reasoning to determine that $(x-y)^2+(y-z)^4+(z-x)^6$ is a factor? $\endgroup$ – flawr Mar 4 '18 at 9:00
  • $\begingroup$ @flawr thnx. that seems true. then how do I extract $(x-y)^2+(y-z)^2+(z-x)^2$ in the first place ? $\endgroup$ – ss1729 Mar 4 '18 at 9:08
  • $\begingroup$ You could always expand the determinant and then factor it, which should not be difficult as you know the result. $\endgroup$ – flawr Mar 4 '18 at 9:10
  • $\begingroup$ @flawr Ofcause i knw how to do that. i'm trying to do it using factor theorem given the fact that the determinant is a symmetric polynomial of degree 6. $\endgroup$ – ss1729 Mar 4 '18 at 9:12
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    $\begingroup$ Does it help to swap rows 2 and 3, then note it's a circulant, and so the determinant is $-(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2 +c\omega)$ where $a=yz-x^2$ etc and $\omega$ is the cube root of unity? $\endgroup$ – ancientmathematician Mar 4 '18 at 9:27
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Hint: Note that your matrix has the form

$$\pmatrix{a&b&c\cr b&c&a\cr c&a&b\cr }$$

Which has the determinant

$$3abc-c^3-b^3-a^3$$

which can again be factored into

$$-\left(a+b+c\right)\,\left(a^2+b^2+c^2-ab-bc-ac\right)$$

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By taking elementary operation:$$\Delta=\begin{vmatrix} yz-x^2&zx-y^2&xy-z^2\\ zx-y^2&xy-z^2&yz-x^2\\ xy-z^2&yz-x^2&zx-y^2 \end{vmatrix}=\begin{vmatrix} xy+yz+zx-x^2-y^2-z^2&zx-y^2&xy-z^2\\ xy+yz+zx-x^2-y^2-z^2&xy-z^2&yz-x^2\\ xy+yz+zx-x^2-y^2-z^2&yz-x^2&zx-y^2 \end{vmatrix}\\=-(x^2+y^2+z^2-xy-yz-zx)\begin{vmatrix} 1&zx-y^2&xy-z^2\\ 1&xy-z^2&yz-x^2\\ 1&yz-x^2&zx-y^2 \end{vmatrix}\\=-\frac{1}{2}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]\begin{vmatrix} 0&(x+y+z)(z-y)&(x+y+z)(x-z)\\ 0&(x+y+z)(x-z)&(x+y+z)(x-z)\\ 1&yz-x^2&zx-y^2 \end{vmatrix}\\=-\frac{1}{2}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]\begin{vmatrix} (z-y)&(x-z)\\ (x-z)&(x-z)\\ \end{vmatrix}=\frac{1}{2}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2, $$ as we know $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]$.

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  • $\begingroup$ pls read the post. i know this. i am not asking for solving it by usual matrix operations. $\endgroup$ – ss1729 Mar 4 '18 at 20:51
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Note that $$ M:= \begin{bmatrix} yz-x^2&zx-y^2&xy-z^2\\ zx-y^2&xy-z^2&yz-x^2\\ xy-z^2&yz-x^2&zx-y^2 \end{bmatrix} $$ is the matrix of $2\times 2$ cofactors of the matrix: $$ N:= \begin{bmatrix} x & y & z\\ y & z & x\\ z& x & y \\ \end{bmatrix}. $$

Then as usual $MN=(\det N) I$, so that $\det M \det N =(\det N)^3$. As $\det N\not=0$ (as a polynomial) we have that $\Delta =\det M= (\det N)^2$ is the perfect square of a polynomial of degree $3$ -- which is what was asked.

But from this we can see everything: there is clearly a factor $(x+y+z)$ in $\det N$ and a factor $\frac{1}{2}( (x-y)^2 +(y-z)^2 +(z-x)^2))$ in $\det M$. Establishing the value of the constant is trivial.

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