Spivak's Calculus: chapter 2, problem 18(c)

Let $a=\sqrt{2}+2^{\frac{1}{3}}.$ Prove by contradiction. Suppose the contrary that $a$ is rational. Observe that \begin{eqnarray*} 2 & = & (2^{\frac{1}{3}})^{3}\\ & = & (a-\sqrt{2})^{3}\\ & = & a^{3}-3a^{2}\sqrt{2}+3a\cdot2-2^{\frac{3}{2}}\\ & = & a^{3}+6a-\sqrt{2}(3a^{2}+2). \end{eqnarray*} Therefore, \begin{eqnarray*} \sqrt{2} & = & \frac{a^{3}+6a-2}{3a^{2}+2}\in\mathbb{Q}. \end{eqnarray*} It is well-known that $\sqrt{2}$ is irrational (Do I need to prove this fact?). Hence, we arrive a contradiction.