Answer
$\frac{(x+y)(x-y)}{x^2+xy+y^2}$.
Work Step by Step
The given expression is
$=\frac{(x-y)^3}{x^3-y^3}\div \frac{x^2-2xy+y^2}{x^2-y^2}$
Replace divide sign with multiplication sign and reciprocal the right hand side fraction as shown below.
$=\frac{(x-y)^3}{x^3-y^3}\times \frac{x^2-y^2}{x^2-2xy+y^2}$
Factor each denominator and numerator.
$=(x-y)^3$
In the factor form.
$=(x-y)(x-y)(x-y)$
$=x^3-y^3$
Use algebraic identity $a^3-b^3=(a-b)(a^2+ab+b^2)$.
$=(x-y)(x^2+xy+y^2)$
$=x^2-y^2$
Use algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x+y)(x-y)$
$=x^2-2xy+y^2$
Rewrite middle term.
$=x^2-xy-xy+y^2$
Group terms.
$=(x^2-xy)+(-xy+y^2)$
Factor from each group.
$=x(x-y)-y(x-y)$
Factor out common term.
$=(x-y)(x-y)$
Substitute all factors into the given expression.
$=\frac{(x-y)(x-y)(x-y)}{(x-y)(x^2+xy+y^2)}\times \frac{(x+y)(x-y)}{(x-y)(x-y)}$
Cancel common factors.
The remaining fraction is
$=\frac{(x+y)(x-y)}{x^2+xy+y^2}$.
