Divisibility: $60 \mid (2x-y)(2y-z)(3z+2x)$, if $8x-10y+27z=0$ As the title says, given $x,y,z \in \mathbb{Z}$, where $8x-10y+27z=0$, prove that $(2x-y)(2y-z)(3z+2x)$ is divisible by $60$.
I tried to bring the formula in a format of $(\cdots)(8x-10y+27z) + 60(\cdots)$, didn't get anywhere.

$z$ is even, so $2$ divides $2 y - z$ and $3 z + 2 x$.
Modulo $3$ you have $0 \equiv 8 x - 10 y + 27 z \equiv 2 x - y$, so $3$ divides $2 x - y$.
Modulo $5$ you have $0 = -8x - 27 z \equiv 2 x + 3 z$, so $3 z + 2 x$ is divisible by $5$.
