Question

Помогите решить сложный пример.

Answer 1

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