calculus - Solving the Riemann Sum $sum_{i=1}^{n}(1+frac{6i}{n})^3(frac{2}{n})$?

So I have the Riemann sum. $\sum_{i=1}^{n}(1+\frac{6i}{n})^3(\frac{2}{n})$. From my understanding that turns into $(\frac{2}{n})\sum_{i=1}^{n}(1+\frac{6i}{n})^3$ and what is really perplexing me is that that summation turns into
$$
\left(\frac{2}{n}\right)\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^3\rightarrow \left(\frac{2}{n^4}\right)\sum_{i=1}^{n}(n+6i)^3$$From there on out I can solve the summation, but I'm having trouble understanding how the $n$ was pulled out and the new summation was formed. If someone could explain that bit I would greatly appreciate it, thanks in advance!

Answer

$$\frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^{3}=\frac{n^{3}}{n^{3}}\frac{2}{n}\sum_{i=1}^{n}\left(1+\frac{6i}{n}\right)^{3}=\frac{2}{n\cdot n^{3}}\sum_{i=1}^{n}n^{3}\left(1+\frac{6i}{n}\right)^{3}$$

and $$n^{3}\left(1+\frac{6i}{n}\right)^{3}=\left(n+\frac{6in}{n}\right)^{3}=\left(n+6i\right)^{3}$$