Question

The value of the expression $\frac{{\left(x^2{-y}^2\right)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3}{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}$ is

• A. $(x^2-y^2)(y^2-z^2)(z^2-x^2)$
• B. $3(x-y)(y-z)(z-x)$
• C. $(x+y)(y+z)(z+x)$
• D. $3(z+y)(y+z)(z+x)$

Answer 1

The correct option is C $(x+y)(y+z)(z+x)$

Given, $\frac{{\left(x^2{-y}^2\right)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3}{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}$

we have $x^2-y^2+y^2-z^2+z^2-x^2=0$

$\because {(x^2-y^2)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$

Similarly , ${(x-y)}^3+{(y-z)}^3+{(z-x)}^3=3(x-y)(y-z)(z-x)$

So, the given expression is

$\frac{{\left(x^2{-y}^2\right)}^3+{(y^2-z^2)}^3+{(z^2-x^2)}^3}{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}$

$\Rightarrow \frac{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}{3(x-y)(y-z)(z-x)}$

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

$\Rightarrow (x+y)(y+z)(z+x)$