<p>For \(x \neq y \neq z\), \(\begin{vmatrix} 1+x^3 & x^2 & 1 \\ 1+y^3 & y^2 & 1 \\ 1+z^3 & z^2 & 1 \end{vmatrix} = 0\) if \(xyz\) is</p>
Step-by-Step Solution
Key Concept: Factor the determinant by decomposing the first column as (1+a³) = 1 + a³, then use column operations to reveal that the determinant equals (x-y)(y-z)(z-x)(xy+yz+zx) after factoring out common terms.
<p><strong>Step 1:</strong> Split the first column using determinant property: C₁ = (1+x³) can be written as 1 + x³</p><p>$$\begin{vmatrix} 1+x^3 & x^2 & 1 \\ 1+y^3 & y^2 & 1 \\ 1+z^3 & z^2 & 1 \end{vmatrix} = \begin{vmatrix} 1 & x^2 & 1 \\ 1 & y^2 & 1 \\ 1 & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x^3 & x^2 & 1 \\ y^3 & y^2 & 1 \\ z^3 & z^2 & 1 \end{vmatrix}$$</p><p><strong>Step 2:</strong> The first determinant is zero (columns 1 and 3 are identical). For the second, factor: $\begin{vmatrix} x^3 & x^2 & 1 \\ y^3 & y^2 & 1 \\ z^3 & z^2 & 1 \end{vmatrix} = xyz\begin{vmatrix} x^2 & x & 1 \\ y^2 & y & 1 \\ z^2 & z & 1 \end{vmatrix}$</p><p><strong>Step 3:</strong> The Vandermonde-type determinant $\begin{vmatrix} x^2 & x & 1 \\ y^2 & y & 1 \\ z^2 & z & 1 \end{vmatrix} = (x-y)(y-z)(z-x)$</p><p><strong>Step 4:</strong> For the original determinant to equal zero with x ≠ y ≠ z (so (x-y)(y-z)(z-x) ≠ 0), we need:</p><p>$$xyz(x-y)(y-z)(z-x) = 0$$</p><p>Therefore: <strong>xyz = 0</strong></p><p>∴ Answer: C</p>
Correct Answer: C