Докажите тождество 1/3b-1 - 27b^3-3b/9b^2+1 (3b/9b^2-6b+1 - 1/9b^2-1)

Решите задачу:

$\frac{1}{3b - 1} - \frac{27b^3 - 3b}{9b^2+1} *( \frac{3b}{9b^2-6b+1}- \frac{1}{9b^2-1} )= \\ \\ = \frac{1}{3b - 1} - \frac{3b((3b)^2-1)}{9b^2+1} *( \frac{3b}{(3b)^2-2*3b*1+1^2}- \frac{1}{(3b)^2-1^2} )= \\ \\ = \frac{1}{3b - 1} - \frac{3b((3b)^2-1^2)}{9b^2+1} *( \frac{3b}{(3b-1)^2}- \frac{1}{(3b-1)(3b+1)} )= \\ \\ = \frac{1}{3b - 1} - \frac{3b(3b-1)(3b+1))}{9b^2+1} * \frac{3b(3b+1)-1(3b-1)}{(3b-1)^2*(3b+1)} = \\ \\ \\$
$=\frac{1}{3b - 1} - \frac{3b(3b-1)(3b+1)}{9b^2+1} * \frac{9b^2+3b-3b+1}{(3b-1)^2(3b+1)} =\frac{1}{3b - 1} - \frac{3b(3b-1)(3b+1)*(9b^2+1)}{(9b^2+1)(3b-1)^2(3b+1)} = \\ \\ = \frac{1}{3b-1} - \frac{3b}{3b-1} = \frac{1-3b}{3b-1} = \frac{-(3b-1)}{3b-1} = \frac{-1}{1} = -1$