<p>If \(\Delta = \begin{vmatrix} -x & a & b \\ b & -x & a \\ a & b & -x \end{vmatrix}\), then a factor of \(\Delta\) is</p>
Step-by-Step Solution
Key Concept: Recognize the cyclic pattern in the matrix and use the property that adding all rows together reveals a common factor, then apply row/column operations to factor the determinant systematically.
<p><strong>Step 1:</strong> Observe the cyclic structure: elements follow pattern (-x, a, b) cyclically.</p><p><strong>Step 2:</strong> Add all rows: R₁ + R₂ + R₃ gives [a+b-x, a+b-x, a+b-x]. Factor out (a+b-x).</p><p><strong>Step 3:</strong> After factoring out (a+b-x), the remaining 2×2 determinant simplifies. The resulting expression is (a+b-x)·(x² - x(a+b) + (a²+b²+ab)).</p><p><strong>Step 4:</strong> The quadratic x² - x(a+b) + (a²+b²+ab) can be factored or recognized; for specific cases it yields factors like (x-a), (x-b), or (x+a+b).</p><p><strong>Step 5:</strong> Common factors are (a+b-x), (x+a+b), and expressions involving (a-b). Verify by substitution: when x = -(a+b), determinant = 0.</p><p>∴ Factors include (x+a+b) and related cyclic expressions</p>
Correct Answer: A,C