# If $\begin{vmatrix}x^n&x^{n+2}&x^{n+3}\\y^n&y^{n+2}&y^{n+3}\\z^n&z^{n+2}&z^{n+3}\end{vmatrix}$ = $(y-z)(z-x)(x-y)(\large\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ then $n$ is equal to

$(a)\;2\qquad(b)\;-2\qquad(c)\;-1\qquad(d)\;1$

## 1 Answer

Given :
$\begin{vmatrix}x^n&x^{n+2}&x^{n+3}\\y^n&y^{n+2}&y^{n+3}\\z^n&z^{n+2}&z^{n+3}\end{vmatrix}=(y-z)(z-x)(x-y)(\large\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
The determinant degree is $n+(n+2)+(n+3)$
$\Rightarrow 3n+5$
and the degree of RHS =2
$\Rightarrow 3n+5=2$
$\Rightarrow n=-1$
Hence (c) is the correct answer.
answered Nov 22, 2013

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