3.1. Solve the system in real numbers

$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=37 \\
x^{2}+x z+z^{2}=28 \\
y^{2}+y z+z^{2}=19
\end{array}\right.
$$
3.1. Solution. Subtract the second equation from the first, and factor the resulting difference. We get $(y - z)(x + y + z) = 9$. From this, we obtain that $y - z = \frac{9}{x + y + z}$. Similarly, subtracting the third equation from the second, we get the equality $x - y = \frac{9}{x + y + z} = y - z$. From this, denoting $y - z = d$, we get that $x = y + d$ and $z = y - d$. Now substitute these new notations into the equation $y - z = \frac{9}{x + y + z}$ and we get that $3 y d = 9$ or $y d = 3$, from which $y^2 d^2 = 9$. Substituting the introduced notations into any of the equations, we get that $3 y^2 + d^2 = 28$. Solving the resulting system for $y^2$ and $d^2$, we get that $y^2 = 9, d^2 = 1$ or $y^2 = \frac{1}{3}, d^2 = 27$. From the equality $y d = 3$, obtained above, we see that $y$ and $d$ have the same sign. From this, we get four solutions: $(5, 4, 3), (-5, -4, -3), \left(\frac{1}{\sqrt{3}} + 3 \sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} - 3 \sqrt{3}\right)$ and $\left(-\left(\frac{1}{\sqrt{3}} + 3 \sqrt{3}\right), -\frac{1}{\sqrt{3}}, -\left(\frac{1}{\sqrt{3}} - 3 \sqrt{3}\right)\right)$. It is clear that all the found solutions are valid.