Determinant Identity: Elegant Solution Please. Can anyone provide a proof of this identity:
$$\begin{vmatrix}
(b+c)^2 &a^2&a^2\\
b^2&(a+c)^2&b^2\\
c^2&c^2&(a+b)^2\\
\end{vmatrix}=2abc(a+b+c)^3 ?$$
Ideally I would like a series of row column operations resulting in the formula.
No brute force calculations. Note that letting $a=0$ we get two rows equal. So $abc$ is a factor. But I have unable to produce this by a simple series of operations, I mean to get one row or column all of whose entries are divisible by $a$. Getting divisibility by $(a+b+c)^2$ is also not hard just subtract one column from the other two. After that however I have been unable to make any further progress. Hopefully someone here is smarter than I.

$$\begin{vmatrix}
(b+c)^2 &a^2&a^2\\
b^2&(a+c)^2&b^2\\
c^2&c^2&(a+b)^2\\
\end{vmatrix}= \begin{vmatrix}
(b+c)^2-a^2 &0&a^2\\
0&(a+c)^2-b^2&b^2\\
c^2-(a+b)^2&c^2 - (a+b)^2&(a+b)^2\\
\end{vmatrix}$$

$$\begin{vmatrix}
(b+c)^2-a^2 &0&a^2\\
0&(a+c)^2-b^2&b^2\\
c^2-(a+b^2)&c^2 - (a+b)^2&(a+b)^2\\
\end{vmatrix} =(a+b+c)^2\begin{vmatrix}
b+c-a &0&a^2\\
0&a+c-b&b^2\\
c-a-b& c-a-b&(a+b)^2\\
\end{vmatrix}$$

$$=(a+b+c)^2\begin{vmatrix}
b+c-a &0&a^2\\
0&a+c-b&b^2\\
-2b& -2a& 2ab\\
\end{vmatrix} (R_3 \to R_3 -(R_1+R_2))$$

$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix}
a(b+c-a) &0&a^2\\
0&b(a+c-b)&b^2\\
-2ab& -2ab& 2ab\\
\end{vmatrix}$$
After this just do $R_{1} \to R_1 + R_3$ and $R_{2} \to R_2 + R_3$
$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix}
ab+ac &a^2&a^2\\
b^2&ba+ac&b^2\\
0& 0& 2ab\\
\end{vmatrix} ={(a+b+c)^2}\begin{vmatrix}
b+c &a&a\\
b&a+c&b\\
0& 0& 2ab\\
\end{vmatrix}$$
$$={(a+b+c)^3}\begin{vmatrix}
1 &a&a\\
1&a+c&b\\
0& 0& 2ab\\
\end{vmatrix} = {(a+b+c)^3}\begin{vmatrix}
1 &a&a\\
0&c&b-a\\
0& 0& 2ab\\
\end{vmatrix}$$
This gives the required value of $2abc(a+b+c)^3$.
