3/(1-а^3)+3/(1+а^3)+6/(1+а^6)+12/(1+а^12)+24/(1+а^24)+48/(1+а^48)=?

Решите задачу:

$\displaystyle\frac{3}{1-a^3}+\frac{3}{1+a^3} +\frac{6}{1+a^6}+\frac{12}{1+a^{12}}+\frac{24}{1+a^{24}}+\frac{48}{1+a^{48}}=\\ \\ \\ =\frac{6}{1-a^6} +\frac{6}{1+a^6}+\frac{12}{1+a^{12}}+\frac{24}{1+a^{24}}+\frac{48}{1+a^{48}}=\\ \\ \\ =\frac{12}{1-a^{12}}+ \frac{12}{1+a^{12}}+\frac{24}{1+a^{24}}+\frac{48}{1+a^{48}}=\\ \\ =\frac{24}{1-a^{24}}+ \frac{24}{1+a^{24}}+\frac{48}{1+a^{48}}=\frac{48}{1-a^{48}} +\frac{48}{1+a^{48}}=\frac{96}{1-a^{96}}$