Помогите, логарифм!!!

Решите задачу:

$27^{1 - \log_36} - 4^{-\log_40,125} = \frac{27}{27^{\log_36}} - 4^{\log_4(0,125)^{-1}} =\\\\=\frac{27}{(3^3)^{\log_36}} - (0,125)^{-1} = \frac{27}{3^{3\log_36}} - (\frac{1}{8})^{-1} = \frac{27}{3^{\log_36^3}} - 8 = \\\\= \frac{3^3}{6^3} - 8 =\frac{3^3}{(2*3)^3} - 8 =\frac{1}{2^3} - 8 = \frac{1}{8} - \frac{64}{8} = \boxed{-\frac{63}{8}}$