2. Task: Calculate $\sqrt[3]{\frac{x}{2015+2016}}$, where $x$ is the harmonic mean of the numbers

$a=\frac{2016+2015}{2016^{2}+2016 \cdot 2015+2015^{2}}$ and $b=\frac{2016-2015}{2016^{2}-2016 \cdot 2015+2015^{2}}$.

The harmonic mean of two positive numbers $a$ and $b$ is the number $c$ such that $\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)$
Solution:

Let's calculate

$\frac{1}{a}+\frac{1}{b}=\frac{2016^{2}+2016 \cdot 2015+2015^{2}}{2016+2015}+\frac{2016^{2}-2016 \cdot 2015+2015^{2}}{2016-2015}=$

$=\frac{(2016-2015)\left(2016^{2}+2016 \cdot 2015+2015^{2}\right)+(2016+2015)\left(2016^{2}-2016 \cdot 2015+2015^{2}\right)}{(2016+2015)(2016-2015)}$

$=\frac{2016^{3}-2015^{3}+2016^{3}+2015^{3}}{(2016+2015)(2016-2015)}=\frac{2 \cdot 2016^{3}}{4031}$, then

$\frac{1}{c}=\frac{1}{2} \cdot \frac{2 \cdot 2016^{3}}{4031}=\frac{2016^{3}}{4031}$, hence $x=\frac{4031}{2016^{3}}$

Finally, $\sqrt[3]{\frac{x}{2015+2016}}=\sqrt[3]{\frac{4031}{(2015+2016) \cdot 2016^{3}}}=\frac{1}{2016}$

Answer: $\frac{1}{2016}$.