Question
Simplify :$\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
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Answer
$\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
If $a+b+c=0$,
then $a^3+b^3+c^3=3 a b c$
Now,
$\mathrm{x}^2-\mathrm{y}^2+\mathrm{y}^2-\mathrm{z}^2+\mathrm{z}^2-\mathrm{x}^2=0$
$\Rightarrow\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3$
$=3\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)\dots..(1)$
And,
$x-y+y-z+z-x=0$
$\Rightarrow(x-y)^3+(y-z)^3+(z-x)^3$
$=3(x-y)(y-z)(z-x)\dots ....(2)$
Now,
$\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
$ =\frac{3\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)}{3(x-y)(y-z)(z-x)}$
$ =(x+y)(y+z)(z+x)\dots...[$From $(1)$ and $(2)]$
If $a+b+c=0$,
then $a^3+b^3+c^3=3 a b c$
Now,
$\mathrm{x}^2-\mathrm{y}^2+\mathrm{y}^2-\mathrm{z}^2+\mathrm{z}^2-\mathrm{x}^2=0$
$\Rightarrow\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3$
$=3\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)\dots..(1)$
And,
$x-y+y-z+z-x=0$
$\Rightarrow(x-y)^3+(y-z)^3+(z-x)^3$
$=3(x-y)(y-z)(z-x)\dots ....(2)$
Now,
$\frac{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
$ =\frac{3\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)}{3(x-y)(y-z)(z-x)}$
$ =(x+y)(y+z)(z+x)\dots...[$From $(1)$ and $(2)]$
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