Factorize $\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$ using factor theorem 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.
 A: 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
$$-\left(a+b+c\right)\,\left(a^2+b^2+c^2-ab-bc-ac\right)$$
A: 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 $x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]$.
A: 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.
