Prove that $\gcd(3^n-2,2^n-3)=1$ if and only if $\gcd(6^n-4,2^n-3)=1$

Question

Prove that $\gcd(3^n-2,2^n-3)=1$ if and only if $\gcd(6^n-4,2^n-3)=1$ where $n$ is a natural number.

I was thinking of using something with the Euclidean algorithm, but I still don't see how to take into account the fact that the gcd is $1$. We know that $3 \nmid 3^n-2$ and both $3^n-2$ and $2^n-3$ are odd, so that may help.

Answer 1

Let $A=3^n-2$ and $B=2^n-3$. If $n\equiv 3\pmod{4}$, $\;5\mid \gcd(A,B)$ by Fermat's little theorem.
We have $AB+3A+2B=6^n-6$ and $2\nmid B,\; 3\nmid A$, hence: $$ \gcd(A,B) = \gcd(6^n\color{red}{-6}, 2^n-3).$$

Answer 2

I'm confused.

If $n = 3$ then $\gcd(3^n - 2,2^n - 3) = \gcd(3^3 -2,2^3 - 3) = \gcd(27-2,8 - 3) = \gcd(25,5) = 5$.

But $\gcd(6^3 -4,2^3 - 3) = \gcd(216-4, 5) = \gcd(212,5) = 1$.

So the statement is not true.

?????