Solve the following system of equations:

$$
\begin{aligned}
x+y+z & =3 \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & =\frac{5}{12} \\
x^{3}+y^{3}+z^{3} & =45
\end{aligned}
$$
Solution. The basic idea of the solution is to find the cubic equation whose roots are precisely $x, y$, and $z$. Let the cubic equation be

$$
a t^{3}+b t^{2}+c t+d=0
$$

We can consider the leading coefficient of the cubic equation to be 1 $(a=1)$, so the Viète's formulas for a cubic equation are:

$$
\begin{aligned}
x+y+z & =-b \\
x y+y z+z x & =c \\
x y z & =-d
\end{aligned}
$$

We immediately see that $b=-3$. By bringing the left side of the second equation to a common denominator:

$$
\frac{5}{12}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{x y+y z+z x}{x y z}=-\frac{c}{d}
$$

Since $x, y, z$ are the roots of the equation, it follows that

$$
\begin{array}{r}
x^{3}-3 x^{2}+c x+d=0 \\
y^{3}-3 y^{2}+c y+d=0 \\
z^{3}-3 z^{2}+c z+d=0
\end{array}
$$

By adding these three equations:

$$
x^{3}+y^{3}+z^{3}-3 x^{2}-3 y^{2}-3 z^{2}+c(x+y+z)+3 d=0
$$

After rearranging and substituting the known values:

$$
\begin{aligned}
& 45+3 c+3 d=3\left(x^{2}+y^{2}+z^{2}\right) \\
& x^{2}+y^{2}+z^{2}=15+c+d
\end{aligned}
$$

On the other hand, from $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$ we have:

$$
9=x^{2}+y^{2}+z^{2}+2 c=15+3 c+d
$$

From the rearrangement of the second equation, we already know that $d=-\frac{12 c}{5}$, so

$$
9=15+3 c-\frac{12}{5} c
$$

From this, by rearranging:

$$
\frac{3}{5} c=-6, \quad c=-10
$$

Substituting back, we finally get $d=24$. We have obtained the cubic equation whose roots are $x, y, z$:

$$
t^{3}-3 t^{2}-10 t+24=0
$$

By substituting $t=2$, we see that this is one of the roots of the equation. Therefore, the factor $(t-2)$ can be factored out of the equation.

$$
\begin{aligned}
t^{3}-2 t^{2}-t^{2}+2 t-12 t+24 & =0 \\
(t-2)\left(t^{2}-t-12\right) & =0
\end{aligned}
$$

By factoring the quadratic term:

$$
(t-2)(t-4)(t+3)=0
$$

The solutions to the system of equations are the triple $(-3, 2, 4)$ and its permutations.