Question

$\frac{4a^{2} - 3a+5 }{ a^{3} -1 } - \frac{1-2a}{a^{2}+a+1 } + \frac{6}{1-a}$

Answer 1

(4a²-3a+5)/(a-1)(a²+a+1) -(1-2a)/(a²+a+1)-6/(a-1)=
=(4a²-3a+5-a+1+2a²-2a-6a²-6a-6)/(a³-1)=-12a/(a³-1)=12a/(1-a³)