Q.(a) Show that $\begin{vmatrix} 2bc-a^2 & c^2 & b^2 \\ c^2 & 2ca-b^2 & a^2 \\ b^2 & a^2 & 2ab-c^2 \end{vmatrix} = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}^2$. OR

Row-reducing the left determinant and using the known factorization of $\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ shows both sides equal the same squared expression.

Step 1 -- the right-hand determinant. Expanding $\Delta=\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}$ along the first row gives $\Delta = a(cb-a^2)-b(b^2-ac)+c(ab-c^2) = 3abc-a^3-b^3-c^3 = -(a^3+b^3+c^3-3abc)$.

Using the standard factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we get $\Delta = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, so $\Delta^2 = (a+b+c)^2(a^2+b^2+c^2-ab-bc-ca)^2$.

Step 2 -- the left-hand determinant. Apply the row operation $R_1\to R_1+R_2+R_3$ to $L=\begin{vmatrix}2bc-a^2&c^2&b^2\\c^2&2ca-b^2&a^2\\b^2&a^2&2ab-c^2\end{vmatrix}$:

New first-row entries: column 1 gives $(2bc-a^2)+c^2+b^2 = b^2+c^2-a^2+2bc = (b+c)^2-a^2 = (b+c-a)(b+c+a)$; similarly column 2 gives $(a+c-b)(a+b+c)$ and column 3 gives $(a+b-c)(a+b+c)$.

So the new row 1 has a common factor of $(a+b+c)$, which can be pulled out of the determinant: $L = (a+b+c)\begin{vmatrix}b+c-a & a+c-b & a+b-c\\ c^2 & 2ca-b^2 & a^2\\ b^2 & a^2 & 2ab-c^2\end{vmatrix}$.

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