Asymptotic behaviour of $\ln( \binom{n^3}{2n^2} )$

Question

Given $\ln( \binom{n^3}{2n^2} )$.

Determining asymptotic behaviour I have got next result:

$$ \ln( \binom{n^3}{2n^2} ) = \ln( \frac{(n^3)!}{(2n^2)! (n^3 - 2n^2)! } ) \approx $$

$$ \approx 2n^2\ln{n} + O(n^2) $$

Is my result right for asymptotic behaviour of given expression?

Answer 1

Maple agrees: in fact it's $$2 n^2 \ln(n) + (2-2\ln(2)) n^2 - 2 n - \ln(n) - \ln(2 \sqrt{\pi}) - 4/3 + O(1/n)$$