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.
Best Answer
$$\begin{vmatrix} (b+c)^2 &a^2&a^2\\ b^2&(a+c)^2&b^2\\ c^2&c^2&(a+b)^2\\ \end{vmatrix}= \begin{vmatrix} (b+c)^2-a^2 &0&a^2\\ 0&(a+c)^2-b^2&b^2\\ c^2-(a+b)^2&c^2 - (a+b)^2&(a+b)^2\\ \end{vmatrix}$$
$$\begin{vmatrix} (b+c)^2-a^2 &0&a^2\\ 0&(a+c)^2-b^2&b^2\\ c^2-(a+b^2)&c^2 - (a+b)^2&(a+b)^2\\ \end{vmatrix} =(a+b+c)^2\begin{vmatrix} b+c-a &0&a^2\\ 0&a+c-b&b^2\\ c-a-b& c-a-b&(a+b)^2\\ \end{vmatrix}$$
$$=(a+b+c)^2\begin{vmatrix} b+c-a &0&a^2\\ 0&a+c-b&b^2\\ -2b& -2a& 2ab\\ \end{vmatrix} (R_3 \to R_3 -(R_1+R_2))$$
$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix} a(b+c-a) &0&a^2\\ 0&b(a+c-b)&b^2\\ -2ab& -2ab& 2ab\\ \end{vmatrix}$$
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 $2abc(a+b+c)^3$.