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UPSC CDS 2 2021 Math Paper

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Question : 68 of 100

Marks: +1, -0

What is

\frac{x^{4}}{(x^{2}-y^{2})(x^{2}-z^{2})}

+\frac{y^{4}}{(y^{2}-x^{2})(y^{2}-z^{2})}

-\frac{z^{4}}{(z^{2}-x^{2})(z^{2}-y^{2})}

equal to?

$-1$
0
1
$x^{2}+y^{2}+z^{2}$

Solution: 👈: Video Solution

Given expression,

\frac{x^{4}}{(x^{2}-y^{2})(x^{2}-z^{2})}

+\frac{y^{4}}{(y^{2}-x^{2})(y^{2}-z^{2})}

+\frac{z^{4}}{(z^{2}-x^{2})(z^{2}-y^{2})}

=\frac{x^{4}}{(x^{2}-y^{2})(x^{2}-z^{2})}

-\frac{y^{4}}{x^{2}-y^{2}}

+\frac{y^{2}-z^{2}}{(x^{2}-z^{2})(y^{2}-z^{2})}

x^{4}(y^{2}-z^{2})-y^{4}(x^{2}-z^{2})

=\frac{+z^{4}(x^{2}-y^{2})}{(x^{2}-y^{2})(x^{2}-z^{2})(y^{2}-z^{2})}

x^{4}(y^{2}-z^{2})-y^{4}x^{2}+y^{4}z^{2}

=\frac{+z^{4}x^{2}-z^{4}y^{2}}{(x^{2}-y^{2})(x^{2}-z^{2})(y^{2}-z^{2})}

x^{4}(y^{2}-z^{2})-y^{4}x^{2}+z^{4}x^{2}

=\frac{+y^{4}z^{2}-z^{4}y^{2}}{(x^{2}-y^{2})(x^{2}-z^{2})(y^{2}-z^{2})}

x^{4}(y^{2}-z^{2})-x^{2}(y^{4}-z^{4})

=\frac{+y^{2}z^{2}(y^{2}-z^{2})}{(x^{2}-y^{2})(x^{2}-z^{2})(y^{2}-z^{2})}

x^{4}(y^{2}-z^{2})-x^{2}(y^{2}-z^{2})

=\frac{(y^{2}+z^{2})+y^{2}z^{2}(y^{2}-z^{2})}{(x^{2}-y^{2})(x^{2}-z^{2})(y^{2}-z^{2})}

(y^{2}-z^{2})[x^{4}-x^{2}y^{2}

=\frac{-x^{2}z^{2}+y^{2}z^{2}]}{(x^{2}-y^{2})(x^{2}-z^{2})(y^{2}-z^{2})}

=\frac{[x^{4}-x^{2}z^{2}-x^{2}y^{2}+y^{2}z^{2}]}{(x^{2}-y^{2})(x^{2}-z^{2})}

=\frac{x^{2}(x^{2}-z^{2})-y^{2}(x^{2}-z^{2})}{(x^{2}-y^{2})(x^{2}-z^{2})}

=\frac{(x^{2}-z^{2})(x^{2}-y^{2})}{(x^{2}-y^{2})(x^{2}-z^{2})}

=1

Hence, the given expression is equal to 1 .

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