Compute:
$$\frac{1-\log_a^{3}{b} }{(\log_a b+\log_b a+1)\log_a\frac{a}{b}}$$
I tried to expand it :
$$\frac{1-\log_a^{3}{b} }{(\log_a b+\log_b a+1)\log_a\frac{a}{b}}$$
$$=\frac{(1-\log_a{b})(\log_a^{2}b+\log_a b+1)}{(\log_a b+\log_b a+1)(1-\log_a{b})}$$
$$=\frac{(\log_a^{2}b+\log_a b+1)}{(\log_a b+\log_b a+1)}$$
But I got nothing.