Prove that if $p$ is a prime such that $p>3$, then $ab^p - ba^p$ is divisible by $6p$

Question

I'm working on this proof and getting a little tripped up. a and b are natural numbers. You can deduce that $ab^p$ and $ba^p$ are odd if a and b are even. Otherwise, $ab^p$ and $ba^p$ are even. In both cases, they are divisible by 2.

Using Fermat's little theorem, $a^p \equiv a mod(p)$ we can make the following conclusion: p divides $a^p$

p divides $a^p - a$

p divides $ba^p -ab$

So p divides $ab^p -ba$

p divides $ab^p - ba - ba^p - ab$

p divided $ab^p - ba^p$

So we know p and 2 divides the quantity, but I'm stuck on how to prove that 3 also divides it so we can say that $2*3*p = 6p$ divides as well.

If $p=3$ then 3 divides $ab^3-ba^3$, but you need that 3 divides $ab^p-ba^p$ for all prime $p>3$.

Does this answer your question? Prove that $6p$ is always a divisor of $ab^{p} - ba^{p}$. — dxiv, Mar 29 '23 at 02:22
Spelling it out: $$ab^p - ba^p = ab\left(b^{p-1} - a^{p-1}\right).$$ Re divisibility by $~p:~$ if $~a,b~$ both rel. prime to $~p,~$ then by Fermat's Little Theorem, you have that $a^{p-1} \equiv 1 \equiv b^{p-1} \pmod{p}.~$ For divisibility by $~3:~$ assuming that neither $~a~$ nor $~b~$ is a multiple of $~3,~$ then you have that both $~a^{p-1}~$ and $~b^{p-1}~$ will be $~\equiv 1 \pmod{3}.~$ This holds because $~(p-1)~$ is even and $~2^{2k} = 4^k \equiv 1 \pmod{3}.$ — user2661923, Mar 29 '23 at 04:21

Prove that if $p$ is a prime such that $p>3$, then $ab^p - ba^p$ is divisible by $6p$

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