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This question already has an answer here:

Our professor asked us this to prove that

$$ \sqrt[3]{2} + \sqrt[3]{4} \notin \Bbb Q. $$

I know how to prove each one separately that it is irrational, but when it comes to summing two irrational numbers its not certain what the result will be.

How do I solve it in a more certain way?

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marked as duplicate by Martin Sleziak, Davide Giraudo, user147263, Ferra, pjs36 Nov 23 '15 at 23:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let us rewrite $$\sqrt[3]2+\sqrt[3]4=\color{blue}{1+\sqrt[3]2+\sqrt[3]4}-1=\color{blue}{\frac{\left(\sqrt[3]2\right)^3-1}{\sqrt[3]2-1}}-1=\frac{1}{\sqrt[3]2-1}-1.$$ If the left side were rational, what could we say about $\sqrt[3]2$ ?

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Hint. If $\sqrt[3]{2}+\sqrt[3]{4}$ rational, so is $1+\sqrt[3]{2}+\sqrt[3]{4}$, and so is $$ \frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{(\sqrt[3]{2})^3-1}{(\sqrt[3]{2})^2+\sqrt[3]{2}+1}=\sqrt[3]{2}-1 $$ and hence so is $\sqrt[3]{2}$.

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    $\begingroup$ I did not downvote, but why is always extremely hard to follow your solutions? I think your answers are not very pedagogical. $\endgroup$ – ILoveMath Oct 27 '14 at 22:47
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    $\begingroup$ Is this particular answer hard to follow? Which part? I could explain. $\endgroup$ – Yiorgos S. Smyrlis Oct 27 '14 at 22:50
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Hint: Find a polynomial with integer coefficients for which $\alpha = \sqrt[3]{2} + \sqrt[3]{4}$ is a root. Then you can use the Rational Roots Theorem.

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    $\begingroup$ A monic polynomial. $\endgroup$ – lhf Oct 27 '14 at 22:56
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    $\begingroup$ @lhf That's not required for the rational roots theorem. $\endgroup$ – zibadawa timmy Oct 28 '14 at 9:15
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Suppose the sum is a rational number $r$, then $\sqrt[3]{4} = -\sqrt[3]{2} + r$, and cubing both sides: $4 = r^3 - 3r^2a + 3ra^2 - 2$, with $a = \sqrt[3]{2}$. Thus: $6 - r^3 = -3r^2a + 3r(r-a) \to a = \dfrac{6-r^3-3r^2}{-3r^2-3r} \in \mathbb{Q}$. But it can be proved that $a = \sqrt[3]{2} \notin \mathbb{Q}$, a contradiction, and we are done.

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  • $\begingroup$ BTW, how could u replace $a^2$ with $r-a$ and still be mathematically right for every a and r ? $\endgroup$ – Firas Ali Abdel Ghani Oct 27 '14 at 23:22
  • $\begingroup$ By the way I defined $a$ and $r$ it works. $\endgroup$ – DeepSea Oct 27 '14 at 23:35
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Write $\sqrt[3]{2} + \sqrt[3]{4} = \alpha + \alpha^2$, with $\alpha=\sqrt[3]{2}$.

Suppose $\alpha + \alpha^2 = q \in \Bbb Q$. Then $\alpha^2 + \alpha^3 = q \alpha$. Since $\alpha^3=2$, we get

$\quad\alpha + \alpha^2 = q$

$\quad\alpha^2 + 2= q \alpha$

Subtracting these, we get $\alpha-2=q-q \alpha$, and so $\alpha=\frac{q+2}{q+1}\in \Bbb Q$, a contradiction since $\alpha$ is irrational.

This argument is essentially dividing the polynomial $x^3-2$ by $x^2+x-q$, getting remainder $(q+1)x-(q+2)$.

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