Solve the following system of equations:

$$
\begin{aligned}
x+y+z & =12 \\
x^{2}+y^{2}+z^{2} & =230 \\
x y & =-15
\end{aligned}
$$
I. Solution: Eliminate $z$ from (2) using (1). After rearranging and dividing by 2,

$$
x^{2} + xy + y^{2} - 12x - 12y = 43
$$

Multiplying by $x^{2}$ and using equation (3), we arrive at

$$
x^{4} - 12x^{3} - 58x^{2} + 180x + 225 = 0
$$

The first two terms of this equation match the first two terms of the square of the expression $x^{2} - 6x$; the last two terms match the last two terms of the square of $6x + 15$ (or $-6x - 15$). All four terms thus appear in the square of the polynomial $x^{2} - 6x - 15$. Therefore, the left side of the equation can be rewritten as:

$$
\left(x^{2} - 6x - 15\right)^{2} - 64x^{2} = \left(x^{2} - 14x - 15\right)\left(x^{2} + 2x - 15\right)
$$

Thus, we get four values for $x$: the roots of

$$
x^{2} - 14x - 15 = 0 \quad \text{and} \quad x^{2} + 2x - 15 = 0
$$

which are $x = 15, -1; 3, -5$. For each $x$, $y$ can be uniquely determined from (3) and $z$ from (1). The possible solutions are

![](https://cdn.mathpix.com/cropped/2024_05_02_4f4d6435c841e57c4a9ag-1.jpg?height=192&width=552&top_left_y=926&top_left_x=773)

These solutions clearly satisfy (2).

Remark: By trial, we quickly find that -1 is a root of equation (4). Furthermore, due to the symmetric roles of $x$ and $y$, the roots of $y$ also satisfy the same equation. Therefore, the value $y = 15$ derived from (3) for $x = -1$ must also be a root of the equation, which it is. This observation can also lead us to the factorization of the equation used in the above solution.

II. Solution: Subtracting (2) from the square of (1), and then subtracting twice (3), and dividing by 2, we get

$$
(x + y)z = -28
$$

Thus, for the unknowns $z$ and $x + y = u$, the equations

$$
\begin{aligned}
& u + z = 12 \\
& uz = -28
\end{aligned}
$$

hold. Therefore, $z$ and $u$ are the roots of the equation

$$
v^{2} - 12v - 28 = 0
$$

Since $v_1 = 14$ and $v_2 = -2$, $x$ and $y$ satisfy

$$
x + y = 14, \quad xy = -15 \quad \text{or} \quad x + y = -2, \quad xy = 15
$$

These lead to the equations

$$
t^{2} - 14t - 15 = 0, \quad \text{or} \quad t^{2} + 2t - 15 = 0
$$

(where we used $x$ instead of $t$ for the unknown), and thus we arrive at the roots found in the previous solution. These solutions satisfy equations (1), (3), and (5). Since (2) follows from these [if we subtract twice (3) and (5) from the square of (1)], they also satisfy the original system of equations.