В 1, доказать тождество Во в 2(в) сделать, как разность кубов или субов кубов

Решите задачу:

$1)\; (a^{-1}-b^{-1})^2=(\frac{1}{a}-\frac{1}{b})^2=\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}=\frac{b^2-2ab+a^2}{a^2b^2}=\frac{(a-b)^2}{a^2b^2}\\\\2)\; (a^{-1}+b^{-1})^2-(a^{-1}-b^{-1})^2=(\frac{1}{a}+\frac{1}{b})^2-(\frac{1}{a}-\frac{1}{b})^2=\\\\=(\frac{1}{a}+\frac{1}{b}-\frac{1}{a}+\frac{1}{b})(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}-\frac{1}{b})=\frac{2}{b}\cdot \frac{2}{a}=\frac{4}{ab}$

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