Find the limit of the following series $x_n$ [closed]

Find $\lim_{n \to \infty}x_n$, where: $$x_n = \sqrt{n^2+1}+\sqrt{4n^2+2n}-\sqrt[3]{27n^3+n^2}.$$ I am completely clueless about what should I do here, every attempt to solve it led me from $\infty - \infty$ to $\infty * 0$, which is even worse. Please, any small hint regarding how should I proceed would be greatly appreciated.

Hint:

You can split in two indeterminate expressions,

$$\sqrt{n^2+1}-\sqrt[3]{n^3+\frac{n^2}{27}}+2\left(\sqrt{n^2+\frac n2}-\sqrt[3]{n^3+\frac{n^2}{27}}\right).$$

Now use

$$a-b=\frac{a^6-b^6}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}$$ on the two limits separately, and notice that all terms in the denominator are asymptotic to $n^5$. There is no need to compute all terms in the numerators, only sixth and fifth degree.

$$\frac{n^6+\cdots-n^6-\dfrac{2n^5}{27}-\cdots}{6n^5}+2\frac{n^6+\dfrac{3n^5}2+\cdots-n^6-\dfrac{2n^5}{27}-\cdots}{6n^5}\to\frac{25}{54}.$$