HINT: Prove there does not exist a rational number that satisfies $x^3=p/q$

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:

$\dfrac{w^3}{v^3}=\dfrac{p}{q}$

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.

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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? –  GovEcon Dec 12 '12 at 5:19
@AndréNicolas Thank you for pointing out that error, I meant to write rational number. –  GovEcon 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

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.

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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.

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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.

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