Question

Find the value of the determinant \(\begin{vmatrix}a & b & c\\ b & c & a\\ c & a & b\end{vmatrix}\).

Hint:

Expand along the first row and match terms to the identity \(a^3+b^3+c^3-3abc\).

Answer 1

Step 1: Choose row operation approach:
Instead of expanding directly, add all three rows into the first row: $R_1 \to R_1+R_2+R_3$.
This is a valid row operation that does not change the value of the determinant.

Step 2: Apply the row operation:
Each entry of row 1 becomes $a+b+c$, so row 1 is now $(a+b+c,\, a+b+c,\, a+b+c)$.
Take $(a+b+c)$ common from row 1:
\[ \Delta = (a+b+c)\begin{vmatrix}1&1&1\\b&c&a\\c&a&b\end{vmatrix} \]

Step 3: Simplify the remaining determinant using column operations:
Apply $C_2 \to C_2-C_1$ and $C_3\to C_3-C_1$, which also does not change the determinant value.
\[ \begin{vmatrix}1&0&0\\b&c-b&a-b\\c&a-c&b-c\end{vmatrix} \]
Expanding along row 1 (only the first entry is nonzero):
\[ = 1\cdot\big[(c-b)(b-c)-(a-b)(a-c)\big] = -(b-c)^2-(a-b)(a-c) \]

Step 4: Expand and collect terms:
Expanding $(a-b)(a-c)=a^2-ac-ab+bc$ and $(b-c)^2=b^2-2bc+c^2$, then combining:
\[ -(b^2-2bc+c^2)-(a^2-ac-ab+bc) = -(a^2+b^2+c^2-ab-bc-ca) \]
So the full determinant is:
\[ \Delta = (a+b+c)\times\left[-(a^2+b^2+c^2-ab-bc-ca)\right] \]

Final Answer:
This matches the earlier factored form obtained by direct expansion. \[ \boxed{\Delta = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)} \]