<p>If \(\begin{vmatrix} x^n & x^{n+2} & x^{n+3} \\ y^n & y^{n+2} & y^{n+3} \\ z^n & z^{n+2} & z^{n+3} \end{vmatrix} = (x-y)(y-z)(z-x)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\), then \(n\) equals</p>
Step-by-Step Solution
Key Concept: Factor out x^n, y^n, z^n from respective rows, then recognize the resulting determinant as a Vandermonde-type form. The RHS constraint determines n by matching powers on both sides.
<p><strong>Step 1:</strong> Factor x^n, y^n, z^n from rows 1, 2, 3 respectively:</p><p>Determinant = x^n·y^n·z^n · \begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix}</p><p><strong>Step 2:</strong> The remaining determinant is Vandermonde-type. Factor as:</p><p>\begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix} = (x^2 - y^2)(y^2 - z^2)(z^2 - x^2)·\dfrac{1}{x^2y^2z^2}</p><p>= (x-y)(x+y)(y-z)(y+z)(z-x)(z+x)·\dfrac{1}{x^2y^2z^2}</p><p><strong>Step 3:</strong> The LHS becomes:</p><p>x^n·y^n·z^n·(x-y)(y-z)(z-x)·\dfrac{(x+y)(y+z)(z+x)}{x^2y^2z^2}</p><p>= (x-y)(y-z)(z-x)·x^{n-2}y^{n-2}z^{n-2}(x+y)(y+z)(z+x)</p><p><strong>Step 4:</strong> RHS = (x-y)(y-z)(z-x)·\dfrac{xy+yz+zx}{xyz} = (x-y)(y-z)(z-x)·x^{-1}y^{-1}z^{-1}(xy+yz+zx)</p><p><strong>Step 5:</strong> For equality, we need the power exponent on xyz to match: n-2 = -1, so n = 1</p><p>∴ Answer: B (n = 1)</p>
Correct Answer: B