Пожалуйста!!! помогите

Решите задачу:

$\frac{(a+2b)^2}{2(a-4b)}-\frac{(a-2b)^2}{3(a-4b)}-\frac{6ab+b^2}{2(a-4b)}=\frac{3(a+2b)^2-2(a-2b)^2-3(6ab+b^2)}{6(a-4b)}= \\ =\frac{3(a^2+4ab+4b^2)-2(a^2-4ab+4b^2)-3(6ab+b^2)}{6(a-4b)}= \\ =\frac{3a^2+12ab+12b^2-2a^2+8ab-8b^2-18ab-3b^2}{6(a-4b)}=\frac{a^2+2ab+b^2}{6(a-4b)}= \frac{(a+b)^2}{6(a-4b)} \\ \frac{(a+b)^2}{6(a-4b)}=\frac{( -\frac{7}{6}+\frac{1}{6})^2}{6(-\frac{7}{6}-4* \frac{1}{6})}=\frac{( -\frac{6}{6})^2}{6(-\frac{11}{6})}= \frac{(-1)^2}{-11}=- \frac{1}{11}$