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The problem is to find $a,b, c \in \mathbb{Q}$ such that

$ \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = \sqrt[3]{\sqrt[3]{2} -1 } $.

My idea: if such representation exists then* $\sqrt[3]{\sqrt[3]{2} -1 } \in \mathbb{Q}(\sqrt[3]{2})$ and therefore the required representation is nothing but

$ \sqrt[3]{\sqrt[3]{2} -1 } = c_1 + c_2\sqrt[3]{2} + c_3 \sqrt[3]{4}, c_i \in \mathbb{Q}$ — decomposition by the standard basis of the extension.

*- Edit: it is not true, I checked by finding eigenvalues of multiplication by $\sqrt[3]{2} -1 $.

But there are two problems:

  1. It’s computationally hard to find the coefficients. It’s probably doable using eigenvalues but I didn’t try it yet.

  2. It’s actually a school problem so no knowledge of field extension is assumed.

Invincible
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1 Answers1

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Let $x^3=2$ so that $x^3-1=1$. Factor the left-hand side from difference of cubes and move the quadratic factor to the right.

$$x-1=\frac 1{1+x+x^2}=\frac 3{3+3x+3x^2}$$

Recalling that $x^3=2$, then we have that

$$x-1=\frac 3{1+3x+3x^2+x^3}=\frac 3{(1+x)^3}=\frac 19\left(\frac 3{1+x}\right)^3$$

In other words

$$\sqrt[3]{x-1}=\frac 3{\sqrt[3]{9}}\frac {1-x+x^2}{1+x^3}=\frac {1-x+x^2}{\sqrt[3]{9}}$$

Substitute $x=\sqrt[3]2$ to get the simplification.

$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}$$

Frank W
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