Q.Show that $\begin{vmatrix} b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c-a & a+b \end{vmatrix} = 8abc$.

Let $|A| = \begin{vmatrix} b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c-a & a+b \end{vmatrix}$.

Step 1. Put $a = 0$: $|A| = \begin{vmatrix} b+c & -c & -b \\ b-c & c & b \\ c-b & c & b \end{vmatrix}$. Expanding along row 1 gives $(b+c)(cb-bc) - (-c)[(b-c)b - b(c-b)] + (-b)[(b-c)c - c(c-b)] = 0 + c\cdot 2b(b-c) - b\cdot 2c(b-c) = 2bc(b-c) - 2bc(b-c) = 0$. So $a$ is a factor of $|A|$.

Step 2. By the same computation with $b=0$ (or $c=0$), $|A|$ again vanishes, so $b$ and $c$ are also factors of $|A|$. Hence $abc$ divides $|A|$.

Step 3. Every entry of $|A|$ is linear (degree 1) in $a,b,c$, so $|A|$ is a homogeneous polynomial of degree 3. Since $abc$ already has degree 3, the remaining factor is a constant $k$: $|A| = k\cdot abc$.

Step 4. Put $a=b=c=1$: the matrix becomes $\begin{vmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{vmatrix} = 8$, and $abc = 1$, so $k\cdot 1 = 8 \Rightarrow k = 8$.

Factor Theorem — vanish-at-zero substitution for each variable + homogeneous-degree matching

• Assuming the determinant is cyclic-symmetric row-to-row and trying the $(a-b)(b-c)(c-a)$ shortcut, which does not apply here.
• Stopping after finding only one of $a,b,c$ as a factor instead of checking all three.
• Arithmetic slips in the $2\times2$ cofactor expansions when $a=0$ (or $b=0$), giving a false non-zero result.