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

Theorem 7.3 (Factor Theorem for determinants). If every entry of a square matrix $A$ is a polynomial in $x$, and $|A|$ vanishes (equals $0$) at $x=a$, then $(x-a)$ is a factor of $|A|$.

This converts a determinant-simplification problem into ordinary polynomial factoring, and is especially powerful when the determinant is a cyclic symmetric form — each row obtained from the one above it by cycling the variables (e.g. $a\to b\to c\to a$).

Working rules:

• If substituting $x=b$ for $a$ makes two rows/columns of $|A|$ identical, then $(a-b)$ is a factor (by Property 4 of determinants + the Factor Theorem).
• If $r$ rows (or columns) coincide at $x=a$ (for an order-$n$ determinant, $n\ge r$), then $(x-a)^{r-1}$ is a factor.
• In a cyclic symmetric determinant, once $(a-b)$ is found to be a factor, $(b-c)$ and $(c-a)$ follow automatically by the cyclic symmetry.
• Let $m$ = (degree of the product of factors found) $-$ (degree of the product of the leading-diagonal entries). Then the remaining factor is:
  • a constant $k$, if $m=0$; …