Solve the following system of equations:

$$
105(x+y)=120(y+z)=168(z+x)=x y z .
$$
# 1. Problem. Solve the following system of equations:

$$
105(x+y)=120(y+z)=168(z+x)=x y z
$$

I. Solution. Any two terms of the given four-term equality sequence must be equal. For example, writing the equality of the first expression with the other three, we get two linear equations and the product of the unknowns appears only in one equation:

$$
\begin{aligned}
& 105(x+y)=120(y+z) \\
& 105(x+y)=168(z+x) \\
& 105(x+y)=x y z
\end{aligned}
$$

(The equality of the right sides follows from these.)

Express $y$ and $z$ in terms of $x$ from the first two equations. Divide (2) by 15, (3) by 21, and subtract the latter from the former. Thus, $z$ will disappear:

$$
7(x+y)-5(x+y)=8(y+z)-8(z+x)=8(y-x)
$$

from which

$$
y=\frac{15 x}{3}
$$

Based on this, from (2)

$$
z=\frac{7}{8}(x+y)-y=\frac{7}{8} \cdot \frac{8 x}{3}-\frac{5 x}{3}=\frac{2 x}{3}
$$

Substituting these into (4) and rearranging,

$$
105 \cdot \frac{8 x}{3}-\frac{10 x^{3}}{9}=0, \quad x\left(7 \cdot 4 \cdot 9-x^{2}\right)=0
$$

and the roots of this are:

$$
x_{1}=0, \quad x_{2}=\sqrt{7 \cdot 4 \cdot 9}=6 \sqrt{7}, \quad x_{3}=-6 \sqrt{7}
$$

The corresponding values of $y$ and $z$ based on (5) and (6) are:

$$
\begin{array}{lll}
y_{1}=0, & y_{2}=10 \sqrt{7}, & y_{3}=-10 \sqrt{7} \\
z_{1}=0, & z_{2}=4 \sqrt{7}, & z_{3}=-4 \sqrt{7}
\end{array}
$$

All three sets of values satisfy the system of equations.

II. Solution. We can also reach the goal simply if we write three independent equations from (1) by equating each of the first three expressions separately with the fourth, and then express all three unknowns in terms of the product $x y z=p$, and first calculate $p$. From the equations mentioned, immediately by division:

$$
\begin{aligned}
x+y & =\frac{p}{105} \\
y+z & =\frac{p}{120} \\
x+z & =\frac{p}{168}
\end{aligned}
$$

Subtracting the third from the sum of the first two,

$$
\begin{aligned}
& 2 y=\frac{p}{105}+\frac{p}{120}-\frac{p}{168}=\frac{p}{840}(8+7-5)=\frac{p}{84} \\
& y=\frac{p}{168}, \quad \text { and similarly } \quad z=\frac{p}{420}, \quad x=\frac{p}{280}
\end{aligned}
$$

Now, by multiplication,

$$
\frac{p}{420} \cdot \frac{p}{168} \cdot \frac{p}{280}=p, \quad p\left(p^{2}-15 \cdot 6 \cdot 10 \cdot 28^{3}\right)=0
$$

from which, with the subtrahend in the parentheses in the form $15 \cdot 4 \cdot 15 \cdot 28^{2} \cdot 2^{2} \cdot 7$,

$$
p_{1}=0 \quad p_{2}=1680 \sqrt{7}, \quad p_{3}=-1680 \sqrt{7}
$$

and thus with the above expressions, we again get the

$$
x, y, z=\left\{\begin{array}{lcc}
0, & 0, & 0 \\
6 \sqrt{7}, & 10 \sqrt{7}, & 4 \sqrt{7} \\
6 \sqrt{7}, & -10 \sqrt{7}, & -4 \sqrt{7}
\end{array}\right.
$$

systems of roots.