Question

If $a=x(y-z),b=y(z-x)$ and $c=z(x-y)$. What is the value of $\frac{xyz}{abc}(\frac{a^3}{x^3}+\frac{b^3}{y^3}+\frac{c^3}{z^3})$ ?

Answer 1

$\frac{a}{x}=-y-z,\frac{b}{y}=z-x$ and $\frac{c}{z}=x-y$

Now $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=y-z+z-x+x-y=0$

$\therefore \frac{a^3}{x^3}+\frac{b^3}{y^3}+\frac{c^3}{z^3}=3\frac{a}{x}.\frac{b}{y}.\frac{c}{z}=\frac{3abc}{xyz}$

$\therefore$ The given expression $\frac{xyz}{abc}\times \frac{3.abc}{xyz}=3$