Q.(a) Using Factor theorem, prove that $\begin{vmatrix}b+c & a & a^2\\ c+a & b & b^2\\ a+b & c & c^2\end{vmatrix}=(a+b+c)(a-b)(b-c)(c-a)$ OR

The determinant vanishes whenever two of $a,b,c$ coincide, which (by the factor theorem) forces $(a-b)$, $(b-c)$, $(c-a)$ to be factors; matching the remaining degree with $(a+b+c)$ and checking one value confirms the identity.

Let $D=\begin{vmatrix}b+c&a&a^2\\c+a&b&b^2\\a+b&c&c^2\end{vmatrix}$.

Step 1 — D vanishes when $a=b$: Put $a=b$. Row 1 becomes $(b+c,\,a,\,a^2)=(a+c,\,a,\,a^2)$ and Row 2 becomes $(c+a,\,b,\,b^2)=(c+a,\,a,\,a^2)$ — the two rows are identical, so $D=0$. By the factor theorem, $(a-b)$ is a factor of D.

Step 2 — D vanishes when $b=c$: By the same argument (Row 2 and Row 3 become identical), $D=0$, so $(b-c)$ is a factor.

Step 3 — D vanishes when $c=a$: Similarly Row 3 and Row 1 become identical, so $D=0$, and $(c-a)$ is a factor.

Step 4 — degree count: Each term in the expansion of D multiplies one entry from column 1 (degree 1 in a,b,c), one from column 2 (degree 1), and one from column 3 (degree 2), so D is a homogeneous polynomial of total degree 4. The three factors found so far, $(a-b)(b-c)(c-a)$, account for degree 3, so the remaining factor must be a homogeneous linear, symmetric-under-cycling expression in $a,b,c$ — i.e. $k(a+b+c)$ for some constant k.

So $D=k(a+b+c)(a-b)(b-c)(c-a)$.

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