Exercise 72Number theory › IrrationalityHigh schoolAdded 3 Nov 2026

Irrationality of a cubic root expression

#cubic root#irrationality#algebraic

Statement

Show that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational.

Hint

Let $\alpha=\sqrt[3]{2}$ . Then $\alpha^3=2$ and the expression is $\alpha+\alpha^2$ . If $\alpha+\alpha^2\in\mathbb{Q}$ , derive a polynomial equation of degree $\le2$ satisfied by $\alpha$ .

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Solution

Setup. Let $\alpha=\sqrt[3]{2}$ , $\beta=\alpha+\alpha^2$ .

Assume $\beta\in\mathbb{Q}$ . Then $\alpha^2+\alpha-\beta=0$ , so $\alpha$ satisfies a quadratic over $\mathbb{Q}$ .

Contradiction. But $\alpha$ has minimal polynomial $X^3-2$ over $\mathbb{Q}$ (irreducible by Eisenstein at $2$ ), of degree $3$ . Contradiction.

Conclusion. $\beta$ is irrational.

Irrationality of a cubic root expression

Statement

Solution

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