Find all real numbers $x, y, z$ such that

$$
x-y+z=2 \quad x^{2}+y^{2}+z^{2}=30 \quad x^{3}-y^{3}+z^{3}=116
$$
The statement suggests symmetric sums, with a few sign changes. To address this, let's set $t = -y$:

We want to solve:

$$
\begin{aligned}
x + z + t & = 2 \\
x^2 + z^2 + t^2 & = 30 \\
x^3 + z^3 + t^3 & = 116
\end{aligned}
$$

We then calculate the elementary symmetric polynomials $\sigma_1, \sigma_2$, and $\sigma_3$ in $x, z$, and $t$:

$$
\begin{gathered}
\sigma_1 = S_1 = 2 \\
\sigma_2 = \frac{1}{2} \left(S_1^2 - S_2\right) = -13
\end{gathered}
$$

and, finally, $S_3 = S_1 S_2 - (S_1 \sigma_2 - 3 \sigma_3)$, so:

$$
\sigma_3 = \frac{1}{3} \left(S_3 - S_1 S_2 + S_1 \sigma_2\right) = \frac{1}{3} (116 - 60 - 26) = 10
$$

Thus, $x, z$, and $t$ are the roots of $P = X^3 - 2X^2 - 13X - 10$. We notice that $-1$ is a root of $P$, which allows us to factorize $P = (X + 1)(X^2 - 3X - 10)$. We can then solve the quadratic equation: the roots of $P$ are $-1, -2$, and $5$, so $\{x, z, t\} = \{-1, -2, 5\}$. Therefore, the possible values of $(x, y, z)$ are $(-1, 2, 5)$, $(-1, -5, -2)$, $(-2, 1, 5)$, $(-2, -5, -1)$, $(5, 1, -2)$, and $(5, 2, -1)$.