Existence of solution to Congruence relation $(x^2-2)(x^2-6)(x^2-3) \equiv 0\pmod p$ I'm taking the final exam in "Number Theory" tomorrow and stuck with:

Prove that  $\,\,\forall p\in\mathbb{Z}_p\,$ the congruence relation: $$(x^2-2)(x^2-6)(x^2-3) \equiv 0\pmod p$$ has a solution.

Here $p$ is an arbitrary prime number.
Any hints? Ideas?

For $p=2,$
$$(x^2-2)(x^2-3)(x^2-6)\equiv x^2(x^2-1)x^2\pmod2$$
But $x^2(x^2-1)x^2$ is divisible by $x(x-1)$ which being a product of two consecutive integers is always divisible by $2$ 
$$(x^2-2)(x^2-3)(x^2-6)\equiv0\pmod2 $$ for all integer $x$
For $p=3,$
$$(x^2-2)(x^2-3)(x^2-6)\equiv (x^2+1)(x^2)x^2\pmod3$$
But $3\nmid(x^2+1),$ so we need $x^4\equiv0\pmod3\iff x\equiv0$
For prime $p>3,$
Using Legendre Symbol for prime $p$, $$\left(\frac6p\right)=\left(\frac3p\right)\cdot\left(\frac2p\right)$$
At least one of $\displaystyle\left(\frac6p\right),\left(\frac3p\right),\left(\frac2p\right)$ must be $=1$
