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.
Best Answer
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.