Q.Prove that $\begin{vmatrix} bc - a^2 & ca - b^2 & ab - c^2 \\ ca - b^2 & ab - c^2 & bc - a^2 \\ ab - c^2 & bc - a^2 & ca - b^2 \end{vmatrix}$ is divisible by $a + b + c$ and find the quotient.

Concept — recognise the circulant. With $p=bc-a^2,\ q=ca-b^2,\ r=ab-c^2$ the determinant is the cyclic array

Sum: $p+q+r=(ab+bc+ca)-(a^2+b^2+c^2)=-(a^2+b^2+c^2-ab-bc-ca)$.

Differences: $p-q=(b-a)(a+b+c)$, and cyclically, so

Multiplying, the two minus signs cancel: …

The idea

Each row is a cyclic shift of $p,q,r$. Such a circulant has the standard value $3pqr-p^3-q^3-r^3$, which factors as $-(p+q+r)(p^2+q^2+r^2-pq-qr-rp)$. We then substitute back in $a,b,c$.

Step 1 — Name the entries

Step 2 — Circulant value

Step 3 — The sum $p+q+r$

Write $S=a^2+b^2+c^2-ab-bc-ca$, so $p+q+r=-S$.

Step 4 — The second factor

Compute one difference:

Cyclically, $q-r=(c-b)(a+b+c)$ and $r-p=(a-c)(a+b+c)$. Then …

Method: Evaluating a Circulant Determinant to Test Divisibility

Use this method whenever a determinant's three rows are cyclic shifts of the same three expressions (a circulant) and you're asked to show it's divisible by some factor and find the quotient.

Steps

Step 1: Name the repeating entries

Give short names (like $p$, $q$, $r$) to the three distinct expressions that appear, shifted cyclically, in each row. This turns a messy-looking determinant into the standard circulant pattern $\begin{vmatrix}p&q&r\\q&r&p\\r&p&q\end{vmatrix}$.

Step 2: Apply the standard circulant identity

A $3\times3$ circulant has the known factored value

which avoids expanding the determinant term by term.

Step 3: Compute the sum $p+q+r$ and the pairwise differences

Substitute the original expressions back in and simplify $p+q+r$ — it usually collapses to a recognisable symmetric expression. Then compute differences like $p-q$; these often factor neatly, revealing a common factor (such as $a+b+c$) shared by every pairwise difference.

Step 4: Rewrite the second bracket using the differences …

Common Mistakes

Mistake 1: Not recognizing the circulant structure and expanding directly

Why it's wrong: substituting $p=bc-a^2$, $q=ca-b^2$, $r=ab-c^2$ back into a raw $3\times3$ expansion in $a,b,c$ produces a huge, error-prone polynomial. The efficient route is to keep working in $p,q,r$ using the standard identity $3pqr-p^3-q^3-r^3=-(p+q+r)(p^2+q^2+r^2-pq-qr-rp)$ and substitute back only at the end.

Mistake 2: Losing the sign when computing $p+q+r$

Why it's wrong: $p+q+r=(ab+bc+ca)-(a^2+b^2+c^2)$, which is $-S$ where $S=a^2+b^2+c^2-ab-bc-ca$ — a student who writes $p+q+r=S$ (dropping the sign flip) carries an incorrect sign through the rest of the derivation, potentially flipping the final answer's sign or making the "divisible by $a+b+c$" claim look false. …