2. $f(-x)=3(-x)^{3}-(-x)=-3 x^{3}+x=-\left(3 x^{3}-x\right)=-f(x)$ $g(-x)=f^{3}(-x)+f\left(\frac{1}{-x}\right)-8(-x)^{3}-\frac{2}{-x}=-f^{3}(x)-f\left(\frac{1}{x}\right)+8 x^{3}+\frac{2}{x}=-g(x)$

Therefore, $g$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a root of the equation $\Rightarrow$ the sum of the roots is zero, if the roots exist

Check $\quad x=1: \quad f(1)=3 \cdot 1^{3}-1=2, \quad f\left(\frac{1}{1}\right)=f(1)=2, \quad$ substitute $\quad$ into the original equation:10=10 - correct $\Rightarrow x=1$ is a root $\Rightarrow$ the roots exist
Answer: 0.

Problem 8 (2nd version).

Find the sum of the roots of the equation $g^{3}(x)-g\left(\frac{1}{x}\right)=5 x^{3}+\frac{1}{x}$, where $g(x)=x^{3}+x$.

## Solution.

Consider the function $f(x)=g^{3}(x)-g\left(\frac{1}{x}\right)-5 x^{3}-\frac{1}{x}$, then the roots of the original equation are the roots of the equation $f(x)=0$.

Investigate the function $f$ for evenness: