Answer:
[tex]x^6+3x^3+3+\dfrac{1}{x^3}[/tex]
Step-by-step explanation:
We can use the pascal triangle to find the coefficients of the expasion of the cube. It holds that
[tex](x^2 + \dfrac{1}{x})^3=(x^2)^3+3(x^2)^2\dfrac{1}{x}+3x^2(\dfrac{1}{x})^2+(\dfrac{1}{x})^3\\\\=x^6+3x^4\dfrac{1}{x}+3x^2\dfrac{1}{x^2}+\dfrac{1}{x^3}\\\\=x^6+3x^3+3+\dfrac{1}{x^3}[/tex]