Factorization

Solve the system of equations:

$$
\begin{aligned}
& 3 x y z-x^{3}-y^{3}-z^{3}=b^{3} \\
& x+y+z=2 b \\
& x^{2}+y^{2}+z^{2}=b^{2}
\end{aligned}
$$
$b^{3}=3 x y z-x^{3}-y^{3}-z^{3}=(x+y+z)\left(x y+y z+x z-x^{2}-y^{2}-z^{2}\right)=2 b\left(x y+y z+x z-x^{2}-y^{2}-z^{2}\right)$ (see problem $\underline{61005}$ g). But the expression in parentheses is non-positive and turns to zero only when $x=y=z$. Therefore, the equation has a solution only when $b=0$. In this case, the last two equations can be written as $z=-x-y$ and $z^{2}=x^{2}+y^{2}$.

| Problem 78198 Topics: [ | ![](https://cdn.mathpix.com/cropped/2024_05_06_62351a5cb646b08ac6fbg-14.jpg?height=110&width=593&top_left_y=2289&top_left_x=465) |
| :---: | :---: |

Squaring the first of them, we get $x y=0$. Therefore, $x=0, z=-y$ or $y=0, z=-x$.

## Answer

For $b=0$: $(0, t,-t),(t, 0,-t)$; for $b \neq 0$ there are no solutions.