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: Understanding the Concept:
We need to expand a 3x3 determinant with rows that are cyclic shifts of \(a,b,c\).
We use direct cofactor expansion along the first row.

Step 2: Key Formula or Approach:
For a determinant \(\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}\), expansion along row 1 gives \(a_1(b_2c_3-b_3c_2)-b_1(a_2c_3-a_3c_2)+c_1(a_2b_3-a_3b_2)\).

Step 3: Detailed Explanation:
Here the matrix is \(\begin{pmatrix}a&b&c\\b&c&a\\c&a&b\end{pmatrix}\), so expanding along row 1:
\[ \Delta = a(c\cdot b - a\cdot a) - b(b\cdot b - a\cdot c) + c(b\cdot a - c\cdot c) \]
\[ = a(bc-a^2) - b(b^2-ac) + c(ab-c^2) \]
\[ = abc - a^3 - b^3 + abc + abc - c^3 \]
\[ = 3abc - a^3 - b^3 - c^3 \]
This matches the standard algebraic identity \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\), so we factor with a negative sign in front.

Step 4: Factor the result:
\[ \Delta = -(a^3+b^3+c^3-3abc) = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \]

Final Answer:
The determinant equals the negative of the standard cubic sum minus three times the product identity. \[ \boxed{\Delta = 3abc-a^3-b^3-c^3 = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)} \]