prove that if p is odd and $p \nmid x$ then $p \nmid (3x^3-x)$ or $p \nmid (7x^3+3x)$

Question

So far, I have only gotten up to the point where I take the contrapositive,

$p \mid (3x^3-x)$ and $p \mid (7x^3+3x) \Rightarrow $ $p$ is even or $p \mid x$

Then using the elimination method

$p \mid (3x^3-x)$ and $p \mid (7x^3+3x)$ and $p$ is odd $\Rightarrow$ $p \mid x$

I tried playing with the hyphotesis but can't seem to get anynthing close to the conclusion.

Answer 1

Since $p|(3x^3-x)$, then $p|(21x^3-7x)$. Similarly, $p|(21x^3+9x)$. Combining these two implies $p|16x$, and since $16=2^4$ and $p$ is odd, we must have $p|x$.