Prove that for $x^3=3$ there isn't rational solution 
Prove that for $x^3=3$ there isn't rational solution

What I did:
Suppose $x= u /v$ is solution
$$\left(\frac u v\right)^3=3$$
Let's take third root from both sides:
$$\frac u v =\sqrt[3]3$$
and $\sqrt[3]3$ is irrationl ,
my problem is this "and $\sqrt[3]3$ is irrationl " is it well known that this number is irrationl? same as $\pi$?
or maybe there is another why to prove it?
 A: No, you cannot just say that $\sqrt[3]{3}$ is irrational; that is just restating that which you are trying to prove.  Adding "it is well known" does not help.
What you need to do is assume $\sqrt[3]{3}$ is rational; then  $\sqrt[3]{3} = \frac{p}{q}$ with $p,q \in \Bbb{Z}^+$ adn there exists such a pair $p,q$ such that g.c.d$(p,q) = 1$.
Then
$$\frac{p^3}{q^3} = 3 \\
p^3 = 3q^3 \\
p^3 = 3k \implies \exists n: p = 3n \\
27n^3 = 3q^3 \\
 q^3 = 9 n^3 = 3 (3n^3) \implies q = 3m
$$
but then both $p$ and $q$ are multiples of $3$, which contradicts the condition that g.c.d$(p,q) = 1$.  
So if $\sqrt[3]{3}$ is rational it cannot be expressed as a reduced fraction, therefore it is irrational.
A: Just start off my assuming $(u,v) = 1$ i.e the fraction is fully reduced. Then it follows that;
 $$u^3 = 3v^3 \Rightarrow 3 \mid u^3 \Rightarrow 3 \mid u \Rightarrow u = 3M$$
Therefore $(3M)^3 = 27M^3 = 3v^3$ and so $9 \mid v^3$ which implies $3 \mid v^3$ and so $(v,u) \not = 1$ which is a contradiction. 
A: You were on the right track,
Let $$\frac{u}{v}=\sqrt[3]3$$ be a rational number,  where $v,u$ are coprime
and $v\neq 0$
$$\implies\frac{u^3}{v^3}=3$$
$$\implies u^3=3v^3$$
As $v$ and $u$ are coprime, their cubes must also be coprime. But this is a contradiction ($u^3=3v^3$) and hence the assumption that $\sqrt[3]3$ is rational was false.
A: As an elementary approach, I personally prefer to prove that given any truly rational number
$$
\frac ab\notin\mathbb Z
$$
we get truly rational powers
$$
\left(\frac ab\right)^n\notin\mathbb Z
$$
for all $n\in\mathbb N$. Namely, if any prime $p$ divides $b$ but not $a$, the same goes for $b^n$ and $a^n$.

The reasons I like this approach are


*

*simplicity

*generality


One can infer from this that an $n$-th root of an integer is either an integer or irrational. So it follows immediately that $\sqrt 2, \sqrt[3]3,\sqrt[5]{17}$, and generally any $n$-th root of an integer that is not a perfect $n$-th power will be irrational.
