<s>[INST] calculate $\lim_{x\to\infty} x + \sqrt[3]{1-x^3}$ So I multiplied by the conjugate and got $$\lim_{x\to\infty} \frac{x^2-(1-x^3)^\frac{2}{3} + x(1-x^3)^\frac{1}{3}-(1-x^3)}{x-(1-x^3)^\frac{2}{3}}$$
and this is where I got stuck.
 [/INST] 
So I multiplied by the conjugate and got

What conjugate expression was that exactly...?
You want to get rid of the cube root by using:
$$a+b=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2-ab+b^2}=\frac{a^3+b^3}{a^2-ab+b^2}$$
with, in your case, $a=x$ and $b=\sqrt[3]{1-x^3}$.
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