Algebra Challenge – 1 ⭐

Solution: Let $y=\sqrt[3]{3}$, we have $$x=y^2+y+1=\dfrac{y^3-1}{y-1}=\dfrac{(\sqrt[3]{3})^3-1}{\sqrt[3]{3}-1}=\dfrac{2}{\sqrt[3]{3}-1}$$ Therefore $$(\dfrac{2}{x}+1)^3=(2\cdot \dfrac{\sqrt[3]{3}-1}{2}+1)^3=(\sqrt[3]{3})^3=\boxed{3}$$