Determine if a real number is rational using roots of polynomial IV

Solution: First, we need to find an equation with integer coefficients such that $\sqrt[3]{5-\sqrt{3}}$ is a solution to it.
Let $x=\sqrt[3]{5-\sqrt{3}}$. Taking cube for both sides, we obtain that $x^3=5-\sqrt{3}$. Hence $x^3-5=-\sqrt{3}$. Squaring both sides, we get
$$
(x^3-5)^2=3,
$$ which is
$$
x^6-10x^3+22=0.
$$ Consider the rational solution of $x^6-10x^2+22=0$. It follows from Corollary 2.3 that those rationa solutions can only be $\pm 1$,$\pm 2$, $\pm 11$, $\pm 22$.
Direct computations would show that none of them will be a solution of $x^6-10x^2+22=0$. Hence $x^6-10x^2+22=0$ has no rational solutions. Therefore, as a solution to $x^6-10x^2+22=0$, $\sqrt[3]{5-\sqrt{3}}$ is irrational.