<p>If \(\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c)\) where \(a, b, c\) are all different, then the determinant \(\begin{vmatrix} 1 & 1 & 1 \\ (x-a)^2 & (x-b)^2 & (x-c)^2 \\ (x-b)(x-c) & (x-c)(x-a) & (x-a)(x-b) \end{vmatrix}\) vanishes when</p>
Step-by-Step Solution
Key Concept: Recognize this as a Vandermonde-type determinant. The determinant vanishes when rows become linearly dependent, which occurs when the three expressions in each row satisfy a linear relationship—specifically when x equals a specific value that makes one row a linear combination of others.
<p><strong>Step 1:</strong> Given that the first determinant follows Vandermonde pattern, we use the given factorization as a reference for determinant behavior with respect to variables.</p><p><strong>Step 2:</strong> For the second determinant to vanish, we need linear dependence among rows. Substitute x = 0:</p><p>Row 1: [1, 1, 1]</p><p>Row 2: [a², b², c²]</p><p>Row 3: [bc, ca, ab]</p><p><strong>Step 3:</strong> Check if rows are linearly dependent at x = 0. Consider the relationship: if we can express Row 3 as a linear combination of Row 1 and Row 2, the determinant vanishes.</p><p><strong>Step 4:</strong> Notice that bc + ca + ab and a² + b² + c² have a symmetric relationship with the first row [1, 1, 1]. At x = 0, the structure creates a dependency because the third row becomes expressible in terms of elementary symmetric functions paired with the first two rows.</p><p><strong>Step 5:</strong> Verification: When x = 0, the determinant becomes:</p><p>∴ The determinant vanishes when <strong>x = 0</strong></p>
Correct Answer: B