What's the algebraic trick to evaluate $\lim_{x\rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$? $$\lim_{x \rightarrow \infty} \frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}$$
I got the first half:
$$\frac{x\sqrt{x}}{\sqrt{x^{3}-1}+x}=\frac{x\sqrt{x}}{\sqrt{x^{3}(1-\frac{1}{x^3})}+x}=\frac{1}{\sqrt{1-\frac{1}{x^3}}+\frac{1}{x^2}}$$
which evaluates to$\frac{1}{1+0}$.
For the second term $\frac{\sqrt[3]{x+1}}{{\sqrt{x^{3}-1}+x}}$ I can't get the manipulation right. Help is much apreciated! 

Note that
$$\frac{x \sqrt{x}+\sqrt[3]{x+1}}{\sqrt{x^{3}-1}+x}=\frac{\sqrt{x^3}}{\sqrt{x^{3}}}\frac{1+\sqrt[6]{\frac{(x+1)^2}{x^9}}}{\sqrt{1-1/x^3}+1/\sqrt x}\to \frac{1+\sqrt{0}}{\sqrt{1-0}+0}$$
