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.

  • 1
    $\begingroup$ 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 $w$? $\endgroup$ – Alex R. Dec 12 '12 at 5:10
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    $\begingroup$ 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? $\endgroup$ – André Nicolas Dec 12 '12 at 5:15
  • $\begingroup$ @Alex Please excuse my mathematical ignorance, but what does $q|v$ stand for? $\endgroup$ – GovEcon Dec 12 '12 at 5:19
  • $\begingroup$ @AndréNicolas Thank you for pointing out that error, I meant to write rational number. $\endgroup$ – GovEcon Dec 12 '12 at 5:19
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    $\begingroup$ $q|v$ means that $q$ divides into $v$ with no remainder. $\endgroup$ – 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.


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.


Problem : Prove generally that a rational fraction in its lowest terms cannot be the cube of a rational number unless p and q are both perfect cubes.

proof: We have, p/q = k^3 let us suppose that x and y are the cube roots of p and q respectively then,

  x^3 = p
  y^3 = q

Now, p/q = k^3 or, p = k^3 . q or, x^3= k^3 . y^3 Hence, x = k.y or,x/y = k

Therefore, this only comes true when x and y are the cube roots of p and q respectively.


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.

  • $\begingroup$ 1 is a perfect cube $\endgroup$ – Divide1918 Dec 14 '20 at 15:05

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