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.
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.
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.