How can I use this theorem

"Let $\alpha$ be a root of the polynomial $x^n+c_{n-1}x^{n-1}+...+c_1x+c_0$, where the coefficients $c_0, c_1, ..., c_n$ are integers. Then $\alpha$ is either an integer or an irrational number."

to show that $\sqrt[3]{5}$ is irrational?

Question

How can I use this theorem

"Let $\alpha$ be a root of the polynomial $x^n+c_{n-1}x^{n-1}+...+c_1x+c_0$, where the coefficients $c_0, c_1, ..., c_n$ are integers. Then $\alpha$ is either an integer or an irrational number."

to show that $\sqrt[3]{5}$ is irrational?

jamesj · Answer

The solutions of $ x^3  -5 $ must therefore be integral or irrational, and it's not hard to show they are not integers.

anonymous · Answer

Ah! I see. I will try that now. Thank you.

anonymous · Answer

Proof:
The solutions of $x^3-5$ must be either integral or irrational. Thus, we are looking for an integer $x$ such that $x^3=5$. However, there is no such integer since $1^3=1<5$ and $2^3=8>5$. Therefore, $x$ must be irrational, and $\sqrt[3]{5}$ is irrational. $\blacksquare$

Is this okay?

jamesj · Answer

Yes

anonymous · Answer

Thank you again.