\[\frac{1}{3b - 1} - \frac{27b^{3} - 3b}{9b^{2} + 1} \cdot \left( \frac{3b}{9b^{2} - 6b + 1} - \frac{1}{9b^{2} - 1} \right) = - 1\]
\[Преобразуем\ левую\ часть\ равенства:\]
\[1)\frac{3b^{\backslash 3b + 1}}{(3b - 1)²} - \frac{1^{\backslash 3b - 1}}{(3b - 1)(3b + 1)} =\]
\[= \frac{9b^{2} + 3b - 3b + 1}{(3b - 1)²(3b + 1)} = \frac{9b^{2} + 1}{(3b - 1)²(3b + 1)}\]
\[2)\ \frac{3b\left( 9b^{2} - 1 \right)}{9b^{2} + 1} \cdot \frac{9b^{2} + 1}{(3b - 1)^{2}(3b + 1)} =\]
\[= \frac{3b(9b^{2} - 1)}{(9b^{2} - 1)(3b - 1)} = \frac{3b}{3b - 1}\]
\[3)\frac{1}{3b - 1} - \frac{3b}{3b - 1} = \frac{1 - 3b}{3b - 1} = - 1\]
\[- 1 = - 1\]
\[Что\ и\ требовалось\ доказать.\]