Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic

Question

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 )

Answer 1

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.

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$$y=x-x^2$$

$$y^2=x^2-2x^3+x^4=x^2-4+2x=x^2+2x-4$$

$$y^3=x^3-3x^4+3x^5-x^6=2-6x+6x^2-4=6x^2-6x-2$$

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.

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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.

Answer 2

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 $$\left(\color{blue}{\sqrt[3]{2}-\sqrt[3]{4}}\right)^3+6\left(\color{blue}{\sqrt[3]{2}-\sqrt[3]{4}}\right)+2=0$$

Answer 3

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 $$x^3 + 6x + 2 = 0.$$