Помогите!!! алгебра. преобразование алгебраических выражений!!

Решите задачу:

$\frac{2}{a} -( \frac{a+1}{a^{3}-1}- \frac{1}{a^2+a+1}- \frac{2}{1-a} ): \frac{a^3+a^2+2a}{a^2-1} = \\ \frac{2}{a} -( \frac{a+1}{(a-1)(a^2+a+1)}- \frac{(a-1)}{(a-1)(a^2+a+1)}+ \frac{2}{a-1} ): \frac{a(a^2+a+2)}{a^2-1} = \\ \frac{2}{a} -( \frac{a+1-(a-1)}{(a-1)(a^2+a+1)}+ \frac{2(a^2+a+1)}{(a-1)(a^2+a+1)} )* \frac{a^2-1}{a(a^2+a+2)} = \\ \frac{2}{a} -( \frac{a+1-a+1}{(a-1)(a^2+a+1)}+ \frac{2a^2+2a+2}{(a-1)(a^2+a+1)} )* \frac{a^2-1}{a(a^2+a+2)} =$
$\frac{2}{a} - \frac{2+2a^2+2a+2}{(a-1)(a^2+a+1)}* \frac{(a-1)(a+1)}{a(a^2+a+2)} = \frac{2}{a} - \frac{2a^2+2a+4}{(a-1)(a^2+a+1)} * \frac{(a-1)(a+1)}{a(a^2+a+2)} = \\ = \frac{2}{a} - \frac{2(a^2+a+2)}{(a-1)(a^2+a+1)} * \frac{(a-1)(a+1)}{a(a^2+a+2)} = \frac{2}{a} - \frac{2}{(a-1)(a^2+a+1)} * \frac{(a-1)(a+1)}{a} = \\ \frac{2}{a} - \frac{2}{(a^2+a+1)} * \frac{(a+1)}{a} = \frac{2}{a} - \frac{2(a+1)}{a(a^2+a+1)} = \frac{2(a^2+a+1)}{a(a^2+a+1)} - \frac{2(a+1)}{a(a^2+a+1)} = \\$
$= \frac{2a^2+2a+2}{a(a^2+a+1)} - \frac{2a+2}{a(a^2+a+1)} = \frac{2a^2+2a+2-(2a+2)}{a(a^2+a+1)} = \frac{2a^2+2a+2-2a-2}{a(a^2+a+1)} = \\ \frac{2a^2}{a(a^2+a+1)} = \frac{2a}{a^2+a+1}$