HMMT February, 2020

Solution:
Call the three equations (1), (2), (3).

(1)/(2) gives $y = 4z$.

(3) $+$ (1) $-$ (2) gives
$$\left(y^{2} - z^{2}\right)x = 15z^{2}x = 240$$
so $z^{2} x = 16$.

Therefore
$$\begin{aligned} & z(x + 2z)^{2} = x^{2} z + z^{2} y + 4z^{2} x = \frac{81}{5} \\ & z(x - 2z)^{2} = x^{2} z + z^{2} y - 4z^{2} x = \frac{49}{5} \end{aligned}$$
so $\left|\frac{x + 2z}{x - 2z}\right| = \frac{9}{7}$.

Thus either $x = 16z$ or $x = \frac{z}{4}$.

If $x = 16z$, then (1) becomes $1024z^{3} + 16z^{3} = 1040$, so $(x, y, z) = (16, 4, 1)$.

If $x = \frac{z}{4}$, then (1) becomes $\frac{1}{4}z^{3} + 16z^{3} = 1040$, so $(x, y, z) = (1, 16, 4)$.