Proof that $\sqrt[3]{3}$ is irrational Proof that $\sqrt[3]{3}$ (Just need a check)
Let $\sqrt[3]{3}= \frac{a}{b}$ both a,b are integers of course. $\Rightarrow 3=\frac{a^{3}}{b^{3}}$ $\Rightarrow$ $3b^{3}=a^{3}$ $\Rightarrow$ 3b=a $\Rightarrow$ $3b^{3}=27b^{3}$ and this is a contradiction because the cubing function is a 1-1 function. Does this work? 

Supposing $\sqrt[3]{3}$ is a rational solution of an equation, we can assume $x=\sqrt[3]{3}$ and $x^3=3$.
But $x^3=3 \Rightarrow x^3-3 =0$. The rational-root theorem tells us that for a polynomial $f(x) = a_nx^n + a_{n−1}x^{n−1} + \ldots + a_1x + a_0$ all of whose coefficients ($a_n$ through $a_0$) are integers, the real rational roots are always of the form $\frac{p}{q}$, where $p$ is an integer divisor of $a_0$ and $q$ is an integer divisor of $a_n$. In this equation the possible rational roots could be $3,-3,1,-1$ and there wasn't any $\sqrt[3]{3}$. So $\sqrt[3]{3}$ is an irrational number.
