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I know that in most cases you assign the expression to a number alpha and then do some algebra in order to reach a polynomial with rational coeficients that alpha is a root of, proceding to prove that the polynomial has no rational roots. However, I can't do the algebra, as I cannot get rid of the cubic roots. Can you help me?

Prove that $2^{1/3} - \frac{1}{2^{1/3}}$ is irrational

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    Consider the cubic conjugate $2^{2/3}+1+{1\over 2^{2/3}}$ – 2017-01-10

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If $x=2^{1/3}-2^{-1/3}$, then

$$x^3=2-3\cdot2^{2/3}\cdot2^{-1/3}+3\cdot2^{1/3}\cdot2^{-2/3}-2^{-1}={3\over2}-3x$$

so that $x$ satisfies the cubic equation

$$2x^3+6x-3=0$$

which has no rational roots for any number of reasons, including Eisenstein's criterion with $p=3$.

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assume that $$2^{1/3}-\frac{1}{2^{1/3}}=\frac{m}{n}$$ Setting $t=2^{1/3}$ then we have the equation $$t^2-\frac{m}{n}t-1=0$$ can you proceed?

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    Oh I see, thank you. Hadn't occured to me to use a contradiction. Have a nice day sir! – 2017-01-10
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    yea here in Germany comes the night! – 2017-01-10