Question

If \( a, b \) and \( c \) are distinct reals and the determinant \[ \begin{vmatrix} a^3+1 & a^2 & a \\ b^3+1 & b^2 & b \\ c^3+1 & c^2 & c \end{vmatrix} = 0, \] then the product \( abc \) is

• A. \( -1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 2 \)
• E. \( 3 \)

Hint:

When variables are distinct, Vandermonde determinant is non-zero, so focus on the remaining factor.

Answer 1

The Correct Option is A

Concept: A determinant is zero if its rows (or columns) are linearly dependent. For distinct \(a,b,c\), a Vandermonde-type determinant is non-zero, so any factor multiplying it must be zero.

Step 1: Rewrite the determinant. \[ \begin{vmatrix} a^3+1 & a^2 & a \\ b^3+1 & b^2 & b \\ c^3+1 & c^2 & c \end{vmatrix} \]

Step 2: Use column operations. Apply: \[ C_1 \rightarrow C_1 - a \cdot C_2 + a^2 \cdot C_3 \] This simplifies the determinant and factors out: \[ = (abc + 1) \begin{vmatrix} 1 & a^2 & a \\ 1 & b^2 & b \\ 1 & c^2 & c \end{vmatrix} \]

Step 3: Use property of Vandermonde determinant. Since \(a,b,c\) are distinct: \[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \neq 0 \] Thus: \[ abc + 1 = 0 \Rightarrow abc = -1 \]