I can’t solve this limit 
I have tried Taylor series (impossible because it doesn’t work at infinity), multiply by conjugate, and usual operations with limits like limit of sum is sum of limits. I have already know about Laurent series, but this task was suggested to first year students, so I hope it’s possible to solve it without so powerful method.
On
One technique you could try is $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow 0^+}f(1/x)$. Using this technique, the above would become:
$$\lim_{x\rightarrow 0^+}\left((1/x^3+1/x+1)\sin x-\sqrt[3]{1/x^6+1}\right)=\lim_{x\rightarrow 0^+}\left(\frac{(1+x^2+x^3)\sin x-\sqrt[3]{1+x^6}}{x^3}\right)$$
I hope you can take it from there, e.g. using L'Hospital's.
One can still use Taylor series. When $x$ gets large, $\frac{1}{x}$ is small. We can use the substitution $u = \frac{1}{x}$, but I will not.
Rewrite the limit as
$$ (x^3+x+1)\sin\left(\frac{1}{x}\right) - x^2\sqrt[3]{1+\frac{1}{x^6}} \sim (x^3+x+1)\cdot\left(\frac{1}{x}-\frac{1}{6x^3}\right) - x^2$$
$$= \frac{5}{6} + \frac{1}{x} - \frac{1}{6x^2} - \frac{1}{6x^3}$$
which approaches $\frac{5}{6}$ as $x\to \infty$.