Алгебра: Упростить ((a^2+16a+12)/(a^3-8))-((2-3a)/(a^2+2a+4))-(3/(a-2))

Решите задачу:

$\frac{a^2+16a+12}{a^3-8}-\frac{2-3a}{a^2+2a+4}-\frac{3}{a-2}=\frac{a^2+16a+12}{(a-2)(a^2+2+4)}-\frac{2-3a}{a^2+2a+4}-\frac{3}{a-2}=\frac{a^2+16a+12-(2-3a)(a-2)-3(a^2+2a+4)}{(a-2)(a^2+2a+4)}=\frac{a^2+16a+12-2a+4+3a^2-6a-3a^2-6a-12}{(a-2)(a^2+2a+4)}=\frac{a^2+2a+4}{(a-2)(a^2+2a+4)}=\frac{1}{a-2}$