How to get rid of a cubic root in a logarithmic limit? The formula: 
$$\lim_{n \to \infty}\dfrac{\log (1 - n + n^2)}{\log (1 + n + n^{10})^{1/3}}.$$
Thanks for any advice!

Hint:
\begin{eqnarray*}
\frac{\log \left( 1-n+n^{2}\right) }{\log \left( 1+n+n^{10}\right) ^{1/3}}
&=&\frac{\log \left( 1-n+n^{2}\right) }{\frac{1}{3}
\log \left( 1+n+n^{10}\right) } \\
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}
\log \left( n^{10}\left( 1/n^{10}+1/n^{9}+1\right) \right) } \\
&=&\frac{\log n^{2}+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\log
n^{10}+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) }.
\end{eqnarray*}
ADDED. Now that there is already a full answer I complete mine. Using the
rule $\log a^{r}=r\log a$, as commented by David Mitra, and $\log ab=\log
a+\log b$, manipulate algebrically the fraction and rewrite it as
\begin{eqnarray*}
\frac{\log \left( 1-n+n^{2}\right) }{\log \left( 1+n+n^{10}\right) ^{1/3}}
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}%
\log \left( 1+n+n^{10}\right) } \\
&=&\frac{\log \left( n^{2}\left( 1/n^{2}-1/n+1\right) \right) }{\frac{1}{3}%
\log \left( n^{10}\left( 1/n^{10}+1/n^{9}+1\right) \right) } \\
&=&\frac{\log n^{2}+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\log
n^{10}+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) } \\
&=&\frac{2\log n+\log \left( 1/n^{2}-1/n+1\right) }{\frac{1}{3}\times 10\log
n+\frac{1}{3}\log \left( 1/n^{10}+1/n^{9}+1\right) } \\
&=&\frac{2+\frac{\log \left( 1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+%
\frac{1}{3}\frac{\log \left( 1/n^{10}+1/n^{9}+1\right) }{\log n}}.
\end{eqnarray*}
We thus have 
\begin{eqnarray*}
\lim_{n\rightarrow \infty }\frac{\log \left( 1-n+n^{2}\right) }{\log \left(
1+n+n^{10}\right) ^{1/3}} &=&\lim_{n\rightarrow \infty }\frac{2+\frac{\log
\left( 1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+\frac{1}{3}\frac{\log
\left( 1/n^{10}+1/n^{9}+1\right) }{\log n}} \\
&=&\frac{2+\displaystyle\lim_{n\rightarrow \infty }\frac{\log \left(
1/n^{2}-1/n+1\right) }{\log n}}{\frac{10}{3}+\frac{1}{3}\displaystyle%
\lim_{n\rightarrow \infty }\frac{\log \left( 1/n^{10}+1/n^{9}+1\right) }{
\log n}} \\
&=&\frac{2+\frac{\log 1}{\displaystyle\lim_{n\rightarrow \infty }\log n}}{%
\frac{10}{3}+\frac{1}{3}\frac{\log 1}{\displaystyle\lim_{n\rightarrow \infty
}\log n}} \\
&=&\frac{2+0}{\frac{10}{3}+\frac{1}{3}\times 0}=\frac{3}{5}.
\end{eqnarray*}
