Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to request a hint for a problem I am working on form Hardy's a Course of Pure Mathematics.

Question Prove generally that a rational fraction $p/q$ in its lowest terms cannot be the cube of a rational function unless $p$ and $q$ are perfect cubes.

My Work

Claim: $\nexists x\in \mathbb Q$ such that $x^3=\dfrac{p}{q}$ unless $\sqrt[3]{p} , \sqrt[3]{q} \in \mathbb Z$

Attempt: Since $x$ is a rational number, it can be written as the ratio of two integers, $w$ and $v$. $x=w/v$ Therefore, we have:


Since $p$ and $q$ are in lowest possible terms, we can infer that both cannot be even.

Rearranging the above, we get:

$w^3 \times q=v^3 \times p$

Based on this, we can identify two possible cases:

1) $w^3$ and $p$ are even and $v^3$ and $q$ are odd.

2) Vice-verse

My problem is that I cannot seem to carry the proof from here.

Another potential root I was thinking of going was basing a proof on the fact that $p$ and $q$ are integers and then seeing if I can find a contradiction, but I am at a loss on how to begin it this way.

share|improve this question
So you can again assume $gcd(v,w)=1$. Notice that since $gcd(p,q)=1$, then in the equation $w^3q=v^3p$, you MUST have $q|v$ by the fundamental theorem of algebra. So $v=xq$ for some $x$. What can you now say about $w$? –  Alex R. Dec 12 '12 at 5:10
It is not clear what th problem is. It reads "rational function." Rational function with coefficients in what? The beginning of your solution seems to be dealing with rational numbers $p/q$. so which is it, rational functions over some field, or rational numbers? –  André Nicolas Dec 12 '12 at 5:15
@Alex Please excuse my mathematical ignorance, but what does $q|v$ stand for? –  Jordan Mahar Dec 12 '12 at 5:19
@AndréNicolas Thank you for pointing out that error, I meant to write rational number. –  Jordan Mahar Dec 12 '12 at 5:19
$q|v$ means that $q$ divides into $v$ with no remainder. –  Ross Millikan Dec 12 '12 at 5:25

3 Answers 3

up vote 3 down vote accepted

We may assume that $p$ and $q$ are relatively prime. Suppose there are relatively prime integers $a$ and $b$ such that $\left(\dfrac{a}{b}\right)^3=\dfrac{p}{q}$. Then $$a^3q=b^3p.\tag{$1$}$$

It is easy to see that $a^3$ and $b^3$ are relatively prime. For if they are not, then there is a prime $r$ that divides both. But then $r$ divides $a$ and $b$.

Now argue that $a^3$ divides $p$ and $p$ divides $a^3$. Since you asked for a hint, we leave this part out.

Conclude from the above result that $p=\pm a^3$.

A similar argument shows that $q$ is the cube of an ineger.

share|improve this answer

Hint $ $ By the Rational Root Test, $\rm\:(a,b)=1,\ \left(\dfrac{a}b\right)^3\! =\dfrac{p}q\:\Rightarrow\:\begin{eqnarray}\rm a\mid p\\ \rm b\mid q\end{eqnarray}\:\Rightarrow\: \left(\dfrac{a}b\right)^2\! =\dfrac{p/a}{q/b}.\:$ Iterate.

share|improve this answer

What about $x=1, p=1$, and $q=1$? I'm pretty sure $1$ is a rational number, and $1/1$ is in its lowest terms.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.