2. Given

$$
\begin{aligned}
x+y+z & =a \\
x^{2}+y^{2}+z^{2} & =b^{2} \\
x^{-1}+y^{-1}+z^{-1} & =c^{-1}
\end{aligned}
$$

determine $x^{3}+y^{3}+z^{3}$.
2. From the third given equality, we obtain

$$
\frac{x y+y z+z x}{x y z}=\frac{1}{c}
$$

5 points

and from the first two

$$
a^{2}-b^{2}=(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)=2(x y+y z+z x)
$$

5 points

Therefore,

$$
x y z=c \cdot \frac{a^{2}-b^{2}}{2}
$$

5 points

Finally, from

$$
(x+y+z)^{3}=3(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)-2\left(x^{3}+y^{3}+z^{3}\right)+6 x y z
$$

we get

$$
x^{3}+y^{3}+z^{3}=\frac{3 a b^{2}+3 c\left(a^{2}-b^{2}\right)-a^{3}}{2}
$$

10 points