Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic How do I show, step by step, that $\sqrt [3]{2}-\sqrt [3]{4}$ is a root of $x^3+6x+2$? 
Start with $x=\sqrt [3]{2}-\sqrt [3]{4}$ do not use the cubic, the cubic is given for convenience. 
( This is example 4.1.3 from Introductory ANT by Alaca/Williams )

Let $a = 2^{1/3}$ and $b = 4^{1/3} = (2^2)^{1/3} = (2^{1/3})^2 = a^2$.  Then $$(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3a^3(a-b),$$ where upon substituting values we obtain $$(a-b)^3 = 2-4 - 3(2)(a-b) = -2 - 6(a-b).$$  Thus if $x = a-b = 2^{1/3} - 4^{1/3}$, then $$x^3 + 6x + 2 = 0.$$
