Determinant using factor theorem 
Prove
  $$\Delta=\begin{vmatrix}
(y+z)^2 & x^2 & x^2 \\
y^2 & (z+x)^2 & y^2 \\
z^2 & z^2 & (x+y)^2 \\
\end{vmatrix} = 2xyz(x+y+z)^3$$
  using factor theorem.

This is solved in Demonstrate using determinant properties that the determinant of A is equal to $2abc(a+b+c)^3$ using factor theorem.
My Attempt: 
$$
x=0\text{ or }y=0\text{ or }z=0\implies\Delta=0\text{ , So $x,y,z$ are factors of }\Delta.\\
(x+y+z)=0\implies \Delta=\begin{vmatrix}x^2&x^2&x^2\\y^2&y^2&y^2\\z^2&z^2&z^2\end{vmatrix}=0\text{ , So $(x+y+z)$ is a factor of $\Delta$.}
$$
$\color{black}{\text{But how do i extract the remaining term $(x+y+z)^2$ to prove $\Delta=2xyz(x+y+z)^3$ }\color{red}{ ?}}$
Similar Example:
Please check answer of @user348749 in How to solve this determinant,
$$
\Delta'=\begin{vmatrix}
(b+c)^2&ab&ca\\
ab&(a+c)^2&bc\\
ac&bc&(a+b)^2
\end{vmatrix}=2abc(a+b+c)^3
$$
it is said that
$$
(a+b+c)=0\implies\Delta'=\begin{align*}
\begin{vmatrix}
c^2 & ca & bc \\ 
ca & a^2 & ab \\
bc & ab & b^2 \\ 
\end{vmatrix}
=abc\begin{vmatrix}
c & a & b \\ 
c & a & b \\
c & a & b \\ 
\end{vmatrix} 
\end{align*}=0
$$
"Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$."
$\color{black}{\text{How can we say this }\color{red}{ ?}}$
My Understanding:
If the problem was similar to this, answer of @Saibal in Factorise a matrix using the factor theorem,
$$\Delta''=
\begin{vmatrix}
x&y&z\\
x^2&y^2&z^2\\
x^3&y^3&z^3\\
\end{vmatrix}$$
I could without doubt do as below:
$$
x=0\text{ or }y=0\text{ or }z=0\implies\Delta''=0\\
x=y\text{ or }y=z\text{ or }z=x\implies\Delta''=0
$$
Thus, $x,y,z,(x-y),(y-z),(z-x)$ are factors of $\Delta''$. ie. $\Delta''=kxyz(x-y)(y-z)(z-x)$

Your attempt is fine. And all the further explanations you need are already given by user348749 in How to solve this determinant
I could just rephrase this here: Since for $x+y+z=0$, all three columns are identical (same argument as all three rows), $(x+y+z)^2$ is a factor (see user348749's explanation). So the remaining factor is linear.  Since $\Delta$ does not change under $x \leftrightarrow y$ or any other permutation, the remaining linear factor must be symmetric under these exchanges, and only $C \cdot (x+y+z)$ is such a factor. Again, not my arguments so far, but user348749's.
The thing that remains to be done is to calculate the constant. Choose e.g. $x=y=z=1$ and you get 
$$
\Delta=\begin{vmatrix}
4 & 1 & 1 \\
1 & 4 & 1 \\
1 & 1 & 4 \\
\end{vmatrix} = 4\cdot 15 - 2\cdot 3 =54 = C \cdot 3^3 = C \cdot 27
$$
which clearly gives $C =2$.
