elementary number theory - Find $gcd(a,b)$ knowing that $(a^3+6b^2,6480)=216$ and $11a+8b mid 540$

I'm stuck with this problem, it should be simple but I can't get far:

Let $a,b \in \Bbb Z$. Find $\gcd(a,b)$ knowing that $\gcd(a^3+6b^2,6480)=216$ and $11a+8b \mid 540$

From $(a^3+6b^2,6480)=216$ I know that $(\frac{a^3+6b^2}{216},30) = 1$. So $a^3+6b^2$ should have a maximun exponent of $2^3$ and $3^3$ times some other number. I'm not sure on how to combine the data that I have, I just need a hint because there's something that I'm missing or forgetting. Ideas? Thanks!

Answer

The question doesn't ask for $a$ and $b$, but for $\gcd(a,b)$. This turns out to be 6. As folks have pointed out, $a=6a_0$ and $b=6b_0$, giving us $\gcd(a_0^3+b_0^2, 30) = 1$ and $11a_0+8b_0 \mid 90.$ If $d=\gcd(a_0, b_0)$, so that $a_0=kd$ and $b_0=ld$, then this last equation is $\gcd(d^2(k^3d+l^2),30)=1$. Which is to say that $d$ has no common factors with 30. But the divisibility condition says $11kd+8ld \mid 90$, so $d$ must divide 90. The only positive integer dividing 90 but relatively prime to 30 is 1. So $\gcd(a,b) = 6.$