Expand $\left(x^{2}+\frac{3}{x}\right)^{4}, x \neq 0$

Using the binomial theorem,the expansion of $(a+b)^n$ is given by $\sum_{k=0}^{n} {n \choose k} a^{n-k} b^k$.
Here,$a = x^2$,$b = \frac{3}{x}$,and $n = 4$.
$\left(x^2 + \frac{3}{x}\right)^4 = {^4C_0}(x^2)^4 + {^4C_1}(x^2)^3\left(\frac{3}{x}\right) + {^4C_2}(x^2)^2\left(\frac{3}{x}\right)^2 + {^4C_3}(x^2)\left(\frac{3}{x}\right)^3 + {^4C_4}\left(\frac{3}{x}\right)^4$
$= 1 \cdot x^8 + 4 \cdot x^6 \cdot \frac{3}{x} + 6 \cdot x^4 \cdot \frac{9}{x^2} + 4 \cdot x^2 \cdot \frac{27}{x^3} + 1 \cdot \frac{81}{x^4}$
$= x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$