Prove that $\begin{vmatrix} (b+c)^2 & ba & ac\\ ba & (c+a)^2 & cb\\ ca & cb & (a+b)^2 \end{vmatrix}=2abc(a+b+c)^3$
I tried my best for approaching the RHS but it gave no result.
Can anyone please help me out
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Would you be content with a "brute force approach" (i.e. fully expand both sides) ? – Maximilian Janisch Jun 30 '21 at 16:10
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Hint:
\begin{align*} \begin{vmatrix} (b+c)^2 & ba & ac\\ ba & (c+a)^2 & cb\\ ca & cb & (a+b)^2 \end{vmatrix}&=\begin{vmatrix} (b+c)^2+a(b+c) & ba & ac\\ b(c+a)+(c+a)^2 & (c+a)^2 & cb\\ c(a+b)+(a+b)^2 & cb & (a+b)^2 \end{vmatrix}\\ &=(a+b+c)\begin{vmatrix} b+c & ba & ac\\ c+a & (c+a)^2 & cb\\ a+b & cb & (a+b)^2 \end{vmatrix} \end{align*}
$\begin{vmatrix} b+c & ba & ac\\ c+a & (c+a)^2 & cb\\ a+b & cb & (a+b)^2 \end{vmatrix}=\begin{vmatrix} 2a+2b+2c & b(c+a)+(c+a)^2 & c(a+b)+(a+b)^2\\ c+a & (c+a)^2 & cb\\ a+b & cb & (a+b)^2 \end{vmatrix}$
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