Kimse cevaplamadı, cevap benden gelsin:
$\displaystyle \left ( \frac{b\left (2a^3-b^3\right )}{a^3+b^3} \right )^3-\left ( \frac{a\left (2b^3-a^3\right )}{a^3+b^3} \right )^3=\frac{b^3(2a^3-b^3)^3-a^3(2b^3-a^3)^3}{(a^3+b^3)^3}= \\ \displaystyle=\frac{a^{12}-b^{12}+2a^9b^3-2a^3b^9}{(a^3+b^3)^3}=\frac{a^{12}-b^{12}+2a^3b^3(a^6-b^6)}{(a^3+b^3)^3}=\frac{(a^6-b^6)(a^6+b^6+2a^3b^3)}{(a^3+b^3)^3} = \\ \displaystyle=\frac{(a^6-b^6)(a^3+b^3)^2}{(a^3+b^3)^3}=a^3-b^3$