Помогите очен надо уравнени за 8-9 класс надо упростить пример

1) $\frac{(ab^{-1}+a^{-1}b+1) (a^{-1}-b^{-1})^{2} }{a^2b^{-2}+a^{-2}b^2-(ab^{-1}+a^{-1}b)}- \frac{1-a^2}{ab} =$
$= \frac{(ab^{-1}+a^{-1}b+1)(a^{-2}-2a^{-1}b^{-1}+b^{-2})}{a^2b^{-2}+a^{-2}b^2-ab^{-1}-a^{-1}b} - \frac{1-a^2}{ab}=$
$= \frac{a^{-1}b^{-1}+a^{-3}b+a^{-2}-2b^{-2}-2a^{-2}-2a^{-1}b^{-1}+ab^{-3}+a^{-1}b^{-1}+b^{-2}}{a^2b^{-2}+a^{-2}b^2-ab^{-1}-a^{-1}b} - \frac{1-a^2}{ab} =$
$= \frac{a^{-3}b+ab^{-3}-a^{-2}-b^{-2}}{a^2b^{-2}+a^{-2}b^2-ab^{-1}-a^{-1}b} - \frac{1-a^2}{ab}=$
$= \frac{a^{-3}b+ab^{-3}-a^{-2}-b^{-2}}{ab(ab^{-3}+a^{-3}b-b^{-2}-a^{-2})} - \frac{1-a^2}{ab}= \frac{1}{ab} - \frac{1-a^2}{ab}= \frac{1-1+a^2}{ab}= \frac{a^2}{ab}= \frac{a}{b}$

2) $(( \frac{x}{x+y} )^{-2}+ \frac{(x-y)^2+4xy}{y^2+xy} )* \frac{x^6}{x^3y^3-y^6}=$
$= ((\frac{x+y}{x})^2+ \frac{x^2-2xy+y^2+4xy}{y^2+xy})* \frac{x^6}{y^3(x^3-y^3)} =$
$= (\frac{(x+y)^2}{x^2}+ \frac{(x+y)^2}{y(x+y)})*\frac{x^6}{y^3(x^3-y^3)}=$
$=(\frac{x^2+2xy+y^2}{x^2}+ \frac{x+y}{y})*\frac{x^6}{y^3(x^3-y^3)}=$
$= \frac{x^2y+2xy^2+y^3+x^3+x^2y}{x^2y} *\frac{x^6}{y^3(x^3-y^3)}=\frac{(x^3+2x^2y+2xy^2+y^3)x^4}{y^4(x^3-y^3)}=$
$= \frac{x^4(x-y)(x^3+2x^2y+2xy^2+y^3)}{(x-y)(x^3-y^3)y^4}= \frac{x^4(x^4+x^3y-xy^3-y^4)}{y^4(x-y)(x^3-y^3)} = \frac{x^4(x+y)(x^3-y^3)}{y^4(x-y)(x^3-y^3)}=$
$= \frac{x^4(x+y)}{y^4(x-y)}$