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This question already has an answer here:

How do I solve this by prime factorization?

I came across a similar problem on MSE just recently, but I can't find it and I thoroughly searched for it. If anyone can find it, please post it in the comment so that I can delete this question.

Update: I don't know if it makes a difference, but please note that this is cubed root

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marked as duplicate by Thomas Andrews, Davide Giraudo, Emily, Jim, azimut Apr 25 '13 at 22:00

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.

@ThomasAndrews Does it make a difference that this is cubed root? – AlanH Apr 25 '13 at 20:24
Yeah,I missed that. It probably doesn't make much difference, but that particular question is very specific to $\sqrt{3}$, unfortunately - it doesn't include the standard "prime" proof which would work for $\sqrt{3}$ and $\sqrt[3]{3}$. – Thomas Andrews Apr 25 '13 at 20:33
Oops, I mis-read the cubed root, too, when voting duplicate. – Emily Apr 25 '13 at 20:55
@AlanH: The exact same method as used in the $\sqrt{3}$ question will solve your problem here - in fact it shows that $3^{1/n}$ is always irrational as long as $b\geq 2$ is an integer. – Eric Naslund Apr 26 '13 at 1:31
up vote 1 down vote accepted

suppose so then

$$m^3 = 3n^3$$

but if that's true then the prime 3 divides $m$, so write $m = 3m'$ and we have

$$9m'^3 = n^3$$

and so n is a multiple of 3 too, put $n = 3n'$ and we have

$$m'^3 = 3n'^3$$

but this $(m',n')$ pair is smaller than $(m,n)$ so we have infinite descent proving there is no solution

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Let $r=\frac{m}{n}$ s.t $\gcd(m,n)=1$ and $r^3=1$ then we have $m^3=3n^3$ and since $\gcd(m^3,n^3)=1$ then $3$ divides $m^3$ and since $3$ is prime number then $3$ divides $m$ so $m=3m'$ and then we have $9m'^3=n^3$ and $\gcd(m',n)=1$ so $3$ divides $n$ which is absurd.

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Suppose there is such a rational number. We may assume that $r = \frac{m}{n}$ where $\gcd(m,n) = 1$. Whence, $m^3 = 3n^3$ so that $3 \mid m$. Set $m = 3m_1$, the previous equality implies $9m_1^3 = n$ so that $3 \mid n$, a contradiction to the assumption that $\gcd(m,n) = 1$.

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