Find the limit of $x(\sqrt{x^2+1}- \sqrt[3]{x^3+1})$ as $x\to +\infty$. 
Calculate $\lim_{x\to +\infty} x(\sqrt{x^2+1}- \sqrt[3]{x^3+1})$. 

First thing came to my mind is to simplify this to something easier. So multiply the numerator and the denominator by something like we used to do when two square roots involves. But I am trying to find that suitable term but it seems out of my reach. Can anybody help me to solve this ? Any hint or help would be nice . Thanks.

Hint : 
Use that
$$a-b=\frac{a^6-b^6}{(a^2+ab+b^2)(a^3+b^3)}$$
where $a=\sqrt{x^2+1}$ and $b=\sqrt[3]{x^3+1}$ to have
$$\begin{align}\lim_{x\to\infty}x(a-b)&=\lim_{x\to\infty}\frac{(a^6-b^6)x}{(a^2+ab+b^2)(a^3+b^3)}\\\\&=\lim_{x\to\infty}\frac{((x^2+1)^3-(x^3+1)^2)x}{(x^2+1+\sqrt{x^2+1}\sqrt[3]{x^3+1}+(x^3+1)^{2/3})((x^2+1)^{3/2}+x^3+1)}\end{align}$$
Now divide top and bottom by $x^5$.
