Объясните, как решать, пожалуйста.

Решите задачу:

$\frac{( x^{2} + xy+ y^{2})*(x-y) }{(x+y)*(x-y)} = \frac{ x^{3} - y^{3} }{ x^{2} - y^{2} }$
$\frac{3x}{ x^{3} - y^{3} } * \frac{x^{3} - y^{3} }{x^{2} - y^{2} } =\frac{3x}{ x^{2} - y^{2} }$
$\frac{3*(x+y)}{(x-y)*(x+y)} = \frac{3(x+y)}{ x^{2} - y^{2} }$
$\frac{3(x+y)}{ x^{2} - y^{2} }+\frac{3x}{ x^{2} - y^{2} }=\frac{3(2x+y)}{ x^{2} - y^{2} }$
$\frac{3(2x+y)}{ x^{2} - y^{2} }:\frac{2x+y}{ x^{2} +2xy+ y^{2} }=\frac{3(2x+y)}{ x^{2} - y^{2} }* \frac{ x^{2} +2xy+ y^{2} }{2x+y} =\frac{3*( x^{2} +2xy+ y^{2}) }{x^{2} - y^{2}}$
$\frac{3*( x^{2} +2xy+ y^{2}) }{x^{2} - y^{2}}* \frac{3}{x+y} =\frac{9*( x^{2} +2xy+ y^{2}) }{(x-y)*(x+y)*(x+y)}=\frac{9*( x^{2} +2xy+ y^{2}) }{(x-y)*(x^{2} +2xy+ y^{2})}=\frac{9 }{(x-y)}$