Is $7^{6n}-6^{6n}$ always divisible by $13$,$127$ and $559$, for any natural number $n$?

Is $7^{6n}-6^{6n}$ always divisible by $13$,$127$ and $559$, for any natural number $n$?

$7^{6n}-6^{6n}={(7^{3n})}^{2}-{(6^{3n})}^{2}$

${(7^{3n})}^{2}-{(6^{3n})}^{2}=(7^{3n}+6^{3n})(7^{3n}-6^{3n})$

$(7^{3n}+6^{3n})(7^{3n}-6^{3n})=(7^n+6^n)(7^{2n}-7^n\times 6^n+6^{2n})(7^n-6^n)(7^{2n}+7^n\times 6^n+6^{2n})$

when, $n=1$

$7^{6}-6^{6}=(7+6)(7^2-7\times 6+6^2)(7-6)(7^2+7\times 6+6^2)=(13)(43)(127)$

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$7^{6n}-6^{6n}=\left(7^6\right)^n-\left(6^6\right)^n$, and $7^6-6^6=13\cdot43\cdot127=559\cdot127$. Now use the fact that
\begin{align*} a^n-b^n&=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\\ &=\sum_{k=0}^{n-1}a^{n-1-k}b^k\;. \end{align*}
HINT $7^{6n}-6^{6n}= (7^6)^n-(6^6)^n$