$\lim\limits_{x\rightarrow+\infty}\left[(x^3+x^2+1)^{1/3}- (x^2+x+1)^{0.5}\right] = -1/6$? Every method I use seem to get me to something to the extent of $0/0$, stuff I can't work with. Wolfram Alpha claims the answer to this is $-1/6$ but they offer no step by step solution.
Would appreciate any tips and help.

Since $$\alpha-\beta=\frac{\alpha^6-\beta^6}{\alpha^5+...+\beta^5}$$
we have that
\begin{gather}\lim\limits_{x\rightarrow+\infty}({x^3+x^2+1})^{1/3}-{(x^2+x+1)}^{1/2}=\lim\limits_{x\rightarrow+\infty}x[{(1+1/x+1/{x^2})}^{1/3}-{(1+1/x+1/{x^2})^{1/2}}]=
\lim\limits_{x\rightarrow+\infty}x\frac{(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3}{(1+1/x+1/{x^2})^{5/3}+...+{(1+1/x+1/{x^2})}^{5/2}}
\end{gather}
The denominator goes to  $6$ and the numerator:
\begin{gather}\lim\limits_{x\rightarrow+\infty}x[(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3]=\\
\lim\limits_{x\rightarrow+\infty}x(1+1/x+1/{x^2})^2[-1/x-1/{x^2}]=\\
\lim\limits_{x\rightarrow+\infty}-(1+1/x+1/{x^2})^2[1+1/x]=-1
\end{gather}
