Master's Theorem recurrence
$n^{\log_6 8} = n^c$ for some $c&gt;1$ (you can tell that $\log_6 8 &gt; 1$ from the fact that $8&gt;6$). Then $n^c = \omega(n \log n)$.
Indeed:
$$\lim_{n \to +\infty} \frac{n^c}{n \log n} = \lim_{n \to +\infty} \frac{n^{c-1}}{\log n} = \lim_{n \to +\infty} \frac{(c-1)n^{c-2}}{1/n} = \lim_{n \to +\infty} (c-1)n^{c-1} = +\infty.$$