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

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

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

$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix}
a(b+c-a) &0&a^2\\
0&b(a+c-b)&b^2\\
-2ab& -2ab& 2ab\\
\end{vmatrix}$$
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 $2abc(a+b+c)^3$.
A: I appreciate the idea of computing the determinant by using its polynomial properties, and symmetry, thus not computing it really. I will try to go in this direction. Let $F(a,b,c)$ be the function given by the determinant from the OP. We know it is a symmetric polynomial of degree $2+2+2=6$ in $a,b,c$.
(And want to show it is $2abc(a+b+c)^3$.)
Let $x$ be some new transcendental variable, and let us compute (generically)
$$
\begin{aligned}
&F(x-b-c,b,c)
\\
&=
\begin{vmatrix}
(b+c)^2 &(b+c-x)^2&(b+c-x)^2\\
b^2&(x-b)^2&b^2\\
c^2&c^2&(x-c)^2\\
\end{vmatrix}
=
\begin{vmatrix}
2bc & 2(b-x)c&2(c-x)b\\
b^2&(b-x)^2&b^2\\
c^2&c^2&(c-x)^2\\
\end{vmatrix}
\\
&=
2b^3c^3
\begin{vmatrix}
1 & (1-x/b)&(1-x/c)\\
1&(1-x/b)^2&1\\
1&1&(1-x/c)^2\\
\end{vmatrix}
\\
&=
2b^3c^3
\begin{vmatrix}
1 & -x/b&-x/c\\
1&-x/b(2-x/b)&0\\
1&0&-x/c(2-x/c)\\
\end{vmatrix}
\\
&=
2b^2c^2x^2
\begin{vmatrix}
1 & 1& 1\\
1& 2-x/b&0\\
1&0& 2-x/c\\
\end{vmatrix}\ .
\end{aligned}
$$
This is a point where i would like to pause.
Now we have some possibilities.

*

*The first one is further calculating, well, not in the spirit of the OP, but easy to bring to a good end the last determinant. We subtract the first line from the other two. We immediately get $\begin{vmatrix}1-x/b & -1\\-1&1-x/c\end{vmatrix}=(1-x/b)(1-x/c)-1$ and so on.


*But i would like to stop here the calculus, and observe that making $x=0$ in the last determinant we get $\begin{vmatrix}1 & 1& 1\\1 & 2&0\\1&0&2\end{vmatrix}=0$.
(The sum of the last two rows depends linearly on the first row.) So we can finally isolate the factor $x^3$. So we know that $F(a,b,c)$ of homogeneous degree $6$ has the factors $a,b,c$, and $(a+b+c)^3$. So we need only to determine the constant involved, e.g. from $F(-1,1,1)
=\begin{vmatrix}4 & 1& 1\\1 & 0&1\\1&1&0\end{vmatrix}
=\begin{vmatrix}2 & 0& 0\\1 & 0&1\\1&1&0\end{vmatrix}=-2$.
$\square$
