I am attempting to prove the following expression, using elementary row and column operations. I have included the attempted solution.
$$\det \begin{bmatrix} b^2+c^2 & ab & ac \\ ba & c^2 +a^2 & bc\\ ca & cb & a^2+b^2\end{bmatrix}=4a^2b^2c^2$$
Attempted Solution:
$$\begin{matrix} R_{1}\to aR_{1}\\ R_{2}\to bR_{2}\\ R_{3}\to cR_{3}\end{matrix}\mapsto \det\begin{bmatrix}b^2+c^2 & a^2 & a^2\\ b^2 &c^2+a^2 & b^2 \\ c^2 &c^2 &a^2+b^2\end{bmatrix} \tag1$$ $$\begin{matrix}R_{1}\to R_{1}-(R_{2}+R_{3})\mapsto \det\begin{bmatrix}0 &-2c^2 &-2b^2 \\ b^2 &c^2+a^2 &b^2 \\ c^2 & c^2 &a^2+b^2\end{bmatrix}\end{matrix}\tag2$$ $$\begin{matrix}R_{2}\to c^2R_{2}-b^2R_{3}\end{matrix}\mapsto \det\begin{bmatrix}0 &-2c^2&-2b^2\\ 0 & c^4+a^2c^2-b^2c^2 & c^2b^2-a^2b^2-b^4\\ c^2 &c^2 &a^2+b^2\end{bmatrix}\tag3$$
I seem to have made some mistake in the transformation $(3)$. Any pointers as to why this transformation is invalid are appreciated. Thanks.
The transformation you did in step $(3)$ actually involves two steps: $$R_2 \mapsto c^2 R_2 \\ R_2 \mapsto R_2-b^2 R_3$$
So, you have to divide by $c^2$ to preserve the identity of the determinant.