Show that $(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$ 
Let $a,b,c\in\Bbb N$ such that $a>b>c$. Then $K:=(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$.

My attempt : Since each $a,b,c$ are either even or odd, WLOG we may assume $a,b$ are both even or odd. For both cases, $a+b$ and $a-b$ are divisible by $2$ so $K$ is divisible by $4$. Note that any $n\in\Bbb N$ is one of $\overline{0},\overline{1},\overline{2}$ in $\operatorname{mod}3$. Well from this, I can argue anyway but I want to show $K$ is divisible by $3$ more easier or nicer way. Could you help?

Note that $x^2\equiv 0 \quad\textrm{or}\quad 1 \mod 3$ for $x\in\mathbb{N}$.
By Pigeonhole Principle, at least two among $a^2$, $b^2$ and $c^2$ have the same remainder when divided by $3$.
