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UPSC CDS Math Basic operation and Factorisation Questions

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Question : 35 of 77

Marks: +1, -0

What is

\frac{(x-y)(y-z)(z-x)}{(x-y)^3+(y-z)^3+(z-x)^3}

$-\frac{1}{3}$
$\frac{1}{3}$
3
-3

Solution:

If

a+b+c=0

, then

a^3+b^3+c^3=3abc

\therefore\frac{(x-y)(y-z)(z-x)}{(x-y)^3+(y-z)^3+(z-x)^3}

=\frac{(x-y)(y-z)(z-x)}{3(x-y)(y-z)(z-x)}

[(x-y)+(y-z)+(z-x)=0]

=\frac{1}{3}

Alternative Method:

Take x = 1,y = 2, z = 3

We have

\frac{(x-y)(y-z)(z-x)}{(x-y)^3+(y-z)^3+(z-x)^3}

=\frac{(1-2)(2-3)(3-1)}{(1-2)^3+(2-3)^3+(3-1)^3}

=\frac{(-1)\times(-1)\times2}{(-1)^3+(-1)^3+2^3}

=\frac{2}{-1-1+8}=\frac{2}{6}=\frac{1}{3}

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