<s>[INST] Prove $\Delta=\left|\begin{smallmatrix} 3x&-x+y&-x+z\\ -y+x&3y&-y+z\\-z+x&-z+y&3z\end{smallmatrix}\right|=3(x+y+z)(xy+yz+zx)$ 
Prove that$$
\Delta=\begin{vmatrix}
3x&-x+y&-x+z\\
-y+x&3y&-y+z\\
-z+x&-z+y&3z
\end{vmatrix}=3(x+y+z)(xy+yz+zx)
$$
  using factor theorem and polynomials.

My Attempt
$$
\begin{matrix}
\color{red}{3x}&\color{blue}{-x+y}&\color{red}{-x+z}&\color{blue}{3x}&\color{red}{-x+y}\\
-y+x&\color{red}{3y}&\color{blue}{-y+z}&\color{red}{-y+x}&3y\\
\color{red}{-z+x}&\color{blue}{-z+y}&\color{red}{3z}&\color{blue}{-z+x}&\color{red}{-z+y}
\end{matrix}
$$
$\Delta$ is a symmetric polynomial of degree $3$.
If we set $x+y+z=0$ we have $\Delta=0$, thus $(x+y+z)$ is a factor, which I can guess from the column operation $C_1\to C_1+C_2+C_3$.
But, how do I find that the remaining factor is $(xy+yz+zx)$ ?
Of course I'll get $\Delta=0$ if I substitute $(xy+yz+zx)=0$, but how do I guess the term in the first place from the fact that $x+y+z$ is one term and $\Delta$ is a symmetric polynomial ?
Similar Problem
Please check the attempted solution by @David Holden in Demonstrate using determinant properties that the determinant of A is equal to $2abc(a+b+c)^3$,
where he has made the substitution $\Delta(a,b,c)=abc(a+b+c)\left(\lambda(a^2+b^2+c^2)+\mu(ab+bc+ca)\right)$ after obtaining $abc(a+b+c)$ as a factor in order to obtain the remaining factor, which is confusing for me. Is there a better way to guess the remaining factor in my case?
 [/INST] As you have mentioned, $x+y+z$ is a factor. As the determinant is symmetric, the factor other than $x+y+z$ must also be symmetric. As it is quadratic, it must be in the form $a(x^2+y^2+x^2)+b(xy+yz+zx)$. (We don't have to guess. $a(x^2+y^2+x^2)+b(xy+yz+zx)$ is the general form of symmetric quadratic polynomial in $x$, $y$, $z$.)
Put $x=1$, $y=z=0$, we have $a=0$.
Put $x=y=z=1$, we have $b=3$.
</s>