How to show that there is no positive rational number a such that $a^3$ = 2? I'm stuck on this one.
So far, I've tried this:
\begin{align}
a^3 &= 2 \\
a &= \left(\frac mn\right) \\ 
a^3 &= \left(\frac mn\right)^3 = 2 \\
m^3 &= 2n^3 \\
m &= 2p \\
m^3 &= (2p)^3
\end{align}
I'm really confused about what to do after this—the book answer says that $m^3$ becomes $m^3 = 2(4p)^3$. And $2n^3 = 2(4p)^3$. I'm super confused here and can't really understand what's going on here.
Can someone tell what exactly is going on here and how should I prove this?

Assume that $a=p/q$, where $p$ and $q$ are not both even (or make the stronger assumption that $\gcd(p,q)=1$), and that $a^3=2$.  Then $(p/q)^3=2$, or $p^3=2q^3$.  Since the right-hand side is even, the left-hand side must be as well; and since the cube of an odd number is odd, $p$ must be even.  So $p=2p'$ for some integer $p'$.  Now $8(p')^3=2q^3$, or $q^3=4(p')^3$, and hence $q$ must be even as well.  This contradicts the original assumption, completing the proof.
