elementary number theory - 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?

Answer

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.