Решить подробно номер 830 баллов

Решите задачу:

$\frac{ab+b^2}{3} : \frac{b^3}{3a} + \frac{a+b}{b} = \frac{b(a+b)}{3} * \frac{3a}{b^3} + \frac{a+b}{b} = \frac{a(a+b)}{b^2} +\frac{a+b}{b} = \\ \\ \frac{a(a+b)+b(a+b)}{b^2} = \frac{(a+b)^2}{b^2} \\ \\$

$\frac{4xy}{y^2-x^2} :( \frac{1}{y^2-x^2} + \frac{1}{x^2+2xy+y^2} )= \\ \\ \frac{4xy}{(y-x)(y+x)} :( \frac{1}{(y-x)(y+x)} + \frac{1}{(x+y)(x+y)} )= \\ \\ \frac{4xy}{(y-x)(y+x)} : \frac{x+y+y-x}{(y-x)(y+x)(y+x)} = \\ \\ \frac{4xy}{(y-x)(y+x)} * \frac{(y-x)(y+x)(y+x)}{2y} =2x(y+x) \\ \\$