Divisibility: $60 \mid (2x-y)(2y-z)(3z+2x)$, if $8x-10y+27z=0$

Question

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.

Answer 1

$z$ 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 $5$.

Answer 2

$\dfrac{5y-4x}{27}=\dfrac z2=u$(say)

$\implies z=2u, 5y-4x=27u=27u(5-4)\iff5(y-27u)=4(x-27u)$

$\dfrac{4(x-27u)}5=y-27u$ which is an integer As $(4,5)=1,5$ must divide $x-27u,x-27u=5v$ (say)

$\implies x=5v+27u,y=4v+27u$

Can you take it from here?