Q.The value of determinant $\begin{vmatrix} a - b & b + c & a \\ b - c & c + a & b \\ c - a & a + b & c \end{vmatrix}$ is
(A) $a^3 + b^3 + c^3$
(B) $3bc$
(C) $a^3 + b^3 + c^3 - 3abc$
(D) none of these

The key idea is to use row operations to simplify the determinant without expanding it fully.

Step 1: Apply $R_1 \to R_1 + R_2 + R_3$.

The first row becomes:

$(a-b + b-c + c-a,\; b+c + c+a + a+b,\; a+b+c)$

$= (0,\; 2(a+b+c),\; a+b+c)$.

Step 2: Factor $(a+b+c)$ from the first row:

Step 3: Expand along the first row:

Compute the $2\times2$ determinants:

• First: $(b-c)c - b(c-a) = bc - c^2 - bc + ab = ab - c^2$.
• Second: $(b-c)(a+b) - (c+a)(c-a) = (b-c)(a+b) - (c^2 - a^2)$.

Simplify the second:

$(b-c)(a+b) = ab + b^2 - ac - bc$.

So second determinant $= ab + b^2 - ac - bc - c^2 + a^2$

$= a^2 + b^2 - c^2 + ab - ac - bc$.

Now substitute: …

Step 1 — Apply $R_1\to R_1+R_2+R_3$. The new first row is

since column 1 gives $(a-b)+(b-c)+(c-a)=0$. Factoring $(a+b+c)$ from the first row,

Step 2 — Expand along the first row.

The two minors are …

Method: Row/Column Operations to Extract a Common Factor

When every entry of a row (or column) shares a structural pattern that sums to something clean, adding all rows/columns together often factors out a common term and shrinks the determinant to a much easier calculation.

Steps

Step 1: Look for a row/column combination that simplifies

Try an operation such as $R_1 \to R_1 + R_2 + R_3$ (this leaves the determinant's value unchanged) and check whether the resulting row becomes a multiple of a simpler expression in every entry.

Step 2: Factor the common term out

If every entry of the new row shares a factor (e.g. $a+b+c$), pull it out in front of the determinant — this is valid because scaling a row scales the whole determinant by that same factor:

Step 3: Expand the remaining, smaller determinant …

Common Mistakes

Mistake 1: Expanding the $3\times3$ determinant directly instead of spotting the row-sum shortcut

Why it's wrong: Brute-force cofactor expansion of the original entries ($a-b$, $b+c$, etc.) is long and very easy to slip a sign or algebra error in; the problem is specifically constructed so $R_1\to R_1+R_2+R_3$ collapses to a clean common factor. Correct approach: always scan a symmetric-looking determinant for a simplifying row/column operation before expanding term by term.

Mistake 2: Losing a sign or factor while pulling $(a+b+c)$ out of the first row …