Монгол Бодлогын Сан

Заавар: $$(1+2a^{\frac13})^2=1+4a^{\frac13}+4a^{\frac23}$$ $$(8a+1)=(2a^{\frac13}+1)(4a^{\frac23}-2a^{\frac13}+1)$$

Бодолт: 1. $$\dfrac{4a}{(1+2a^{\frac{1}{3}})^3}-\dfrac{a}{8a+1}=\dfrac{4a(4a^{\frac23}-2a^{\frac13}+1)-a(1+4a^{\frac13}+4a^{\frac23})}{(1+2a^{\frac13})^2(8a+1)}=$$ $$=\dfrac{3a(4a^{\frac23}-4a^{\frac13}+1)}{(1+2a^{\frac13})^2(8a+1)}=\dfrac{3a(1-2a^{\frac13})^2}{(1+2a^{\frac13})^2(8a+1)}$$

2. $$\left(\dfrac{1-2a^{\frac{1}{3}}}{1+2a^{\frac{1}{3}}}\right)^2:\dfrac{3a(1-2a^{\frac13})^2}{(1+2a^{\frac13})^2(8a+1)}=$$ $$\left(\dfrac{1-2a^{\frac{1}{3}}}{1+2a^{\frac{1}{3}}}\right)^2\cdot\dfrac{(1+2a^{\frac13})^2(8a+1)}{3a(1-2a^{\frac13})^2}=\dfrac{8a+1}{3a}$$

3. $$\dfrac{1}{3a}-\dfrac{8a+1}{3a}=\dfrac{-8a}{3a}=-\dfrac83$$