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?

It is possible to answer the question by looking directly at modulo $12$.
First, notice that $\forall n,\ n^2 \equiv \{0,1,4,9\} \bmod 12$. So each of $a,b,c$ must have a square with one of those residues. If any two such squares have the same residue, their difference will be $\equiv 0 \bmod 12$, and the product of the differences will be $\equiv 0 \bmod 12$, and we would be done. That leaves to be addressed those cases in which $a^2,b^2,c^2$ all have different residues $\bmod 12$.
There are four ways to select three different residues from four possibilities:
Case 1: Residues of the squares are $9,4,1$. Then the differences are $(9-4)(9-1)(4-1)=5\cdot 8\cdot 3=120 \equiv 0 \bmod 12$
Case 2: Residues of the squares are $9,4,0$. Then the differences are $(9-4)(9-0)(4-0)=5\cdot 9\cdot 4=180 \equiv 0 \bmod 12$
Case 3: Residues of the squares are $9,1,0$. Then the differences are $(9-1)(9-0)(1-0)=8\cdot 9\cdot 1=72 \equiv 0 \bmod 12$
Case 4: Residues of the squares are $4,1,0$. Then the differences are $(4-1)(4-0)(1-0)=3\cdot 4\cdot 1=12 \equiv 0 \bmod 12$
Note that changing the order in which the residues occur in the terms might change the sign of their product, but not its absolute value, and hence not its divisibility by $12$.
