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?

Alternative route:
Let $\alpha^3=2$, and assume $\alpha\in\Bbb Q$, then we have $$\frac{p^3}{q^3}=2\\p^3=2q^3\\p^3=q^3+q^3$$
The last line contradicts Fermat's Last Theorem.


Let $x,y,z$ be non-zero integers. Then if $x^3+y^3=z^3$ we have $xyz=0$.

Proof: We can assume that $x,y,z$ are relatively prime, since if they were not we could divide by $\gcd(x,y,z)^3$.
Reducing the equation modulo $9$, and noticing that the only cubes are $0,-1,1$ we conclude that exactly one of $x,y,z$ is divisible by $9$. Without loss of generality we may assume that $9\mid z$.
We can factor the equation is question as $$z^3=(x+y)(x^2-xy+y^2)$$
Since $9\mid z^3$ we have that $3$ divides one of the factors $x+y, x^2-xy+y^2$. In fact, $3$ divides both of them as $$x^2-xy+y^2=(x+y)^2-3xy$$ That is, $3\mid x+y\iff 3\mid x^2-xy+y^2$. Since $3^3\mid z^3$ one of these factors are divisible by $9$ which will yield a contradiction:

*

*If $9\mid x^2-xy+y^2$ then $9\mid (x+y)^2-3xy$, so $9\mid 3xy$ which is impossible, as $x,y$ are assumed not to be divisible by $3$.


*If $9\mid x+y$, let $k$ be the highest power such that $3^k\mid x+y$, $k\geq 2$. Then we have $$3^{3k}\mid (x+y)^3=x^3+y^3+3xy(x+y)$$ As $3^{3k}\mid x^3+y^3=z^3$, we have $$3^{3k}\mid 3xy(x+y)$$ This is impossible, since then $3^{2k-1}\mid xy$, and $2k-1\geq 3$.
