This is an exercise from R. Courant's book: How to prove $\sqrt3 + \sqrt[3]{2}$ is a irrational number?
The solution is to construct a equation to prove, but is there any other method to prove this, like by contradiction?
Answer
Developing the hint by Winther. Assume it is rational $r=\sqrt{3}+\sqrt[3]{2}$. So we have
$$2=(r-\sqrt{3})^3=r^3-3r^2\sqrt{3}+9r-3\sqrt{3}$$
Regrouping and dividing by $3r^2+3\neq 0$ we get
$$\sqrt{3}={r^3+9r-2\over 3r^2+3}$$
This means $\sqrt{3}$ is rational. Contradiction
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