Question

What is the value of the expression $\frac{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}{(x-y)(y-z)(z-x)}$ ?

Answer 1

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

Numerator:

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

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

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