<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] Here is a "by hand" method.
Note that $\sqrt [3] 4 =(\sqrt [3] 2)^2$ and $(\sqrt [3] 4)^2=2\sqrt[3] 2$.
This means that the powers of $y=\sqrt [3] 2-\sqrt [3] 4$ can all be written as polynomials in $x=\sqrt [3] 2$.
But since $x^3=2$ we can always reduce the polynomial to at most quadratic.

$$y=x-x^2$$
$$y^2=x^2-2x^3+x^4=x^2-4+2x=x^2+2x-4$$
$$y^3=x^3-3x^4+3x^5-x^6=2-6x+6x^2-4=6x^2-6x-2$$
Now eliminate $x^2$ from the final two equations using the first, which tells us that $x^2=x-y$
This gives $$y^2 =x-y+2x-4: \text { or } y^2+y=3x-4$$
and $$y^3=6(x-y)-6x-2=-6y-2: \text { or } y^3+6y+2=0$$
It was possible that we would have had to eliminate $x$ between these two equations to get a cubic in $y$, but we were "lucky" that $x$ disappeared of its own accord. See @Stephanos's answer for why this happens.

We were always going to get expressions for $y, y^2, y^3$ as quadratics in $x$ - i.e. as expressions with a constant term, a term in $x$ and a term in $x^2$. If we treat $\{1, x, x^2\}$ like a basis in linear algebra, with three equations we can systematically eliminate $x$ and $x^2$ to leave a cubic in $y$ (with just the constant term).
Applying the techniques of linear algebra in cases like this is one way in which more general problems of this kind can be tackled - and indeed is a tool in proving powerful general theorems.
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