<s>[INST] 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 )
 [/INST] Raising the given number to the third power and using the identity $$(a-b)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3–3ab(a–b)$$ you obtain $$\left(\sqrt[3]{2}-\sqrt[3]{4}\right)^3=2-4-3\sqrt[3]8\left(\sqrt[3]{2}-\sqrt[3]{4}\right)=-2-6\left(\sqrt[3]{2}-\sqrt[3]{4}\right)$$ or equivalently $$\left(\color{blue}{\sqrt[3]{2}-\sqrt[3]{4}}\right)^3+6\left(\color{blue}{\sqrt[3]{2}-\sqrt[3]{4}}\right)+2=0$$
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