Question

If \[ \left| \begin{array}{ccc} x^{n} & x^{n+2} & x^{n+3} \\ y^{n} & y^{n+2} & y^{n+3} \\ z^{n} & z^{n+2} & z^{n+3} \end{array} \right| =(x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \] then \(n\) is equal to:

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

Hint:

When identities involving variables \(x,y,z\) hold for all values:
Compare total degrees on both sides
Determinants of polynomial form often factor into Vandermonde-type expressions

Answer 1

The Correct Option is D

Step 1: Take \(x^n, y^n, z^n\) common from the rows of the determinant. \[ = x^n y^n z^n \left| \begin{array}{ccc} 1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3} \end{array} \right| \]
Step 2: The standard determinant result: \[ \left| \begin{array}{ccc} 1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3} \end{array} \right| =(x-y)(y-z)(z-x)(x+y+z) \]
Step 3: Hence, LHS becomes: \[ x^n y^n z^n (x-y)(y-z)(z-x)(x+y+z) \]
Step 4: Equating with RHS and cancelling the common factor \((x-y)(y-z)(z-x)\): \[ x^n y^n z^n (x+y+z)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \]
Step 5: Write RHS as: \[ \frac{xy+yz+zx}{xyz} \] \[ \Rightarrow x^{n+1}y^{n+1}z^{n+1}(x+y+z)=xy+yz+zx \]
Step 6: Compare degrees of both sides. LHS degree \(= 3(n+1)+1=3n+4\) RHS degree \(=2\) \[ 3n+4=2 \Rightarrow n=-2 \]