Hey guys I need help with this question

$\displaystyle \text{prove}

(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$

and the answer given in the solution to the question is that since

$\displaystyle 4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$

$\displaystyle 3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$

$\displaystyle \therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$

I know by substituting in values for n in $\displaystyle (1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?

The same goes for$\displaystyle (1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction

If we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?

Thanks