A rational and irrational problem

First we cube both sides of this equation rearranged to get

$\left(a+b \cdot 2^{\frac{1}{3}}\right)^3 =a^3+b^3+3ab\left( a \cdot 2^ { \frac{1}{3} } +b \cdot 2^{ \frac{2}{3} } \right)=-4c$

which implies that if $a, b, c$ are rational, both must be rational also

$\left(a+b \cdot 2^{\frac{1}{3}}\right)$
$\left(a \cdot 2^ { \frac{1}{3} } +b \cdot 2^{ \frac{2}{3} } \right) = 2^{\frac{1}{3}} \left(a+b \cdot 2^{\frac{1}{3}}\right)$

which is an impossibility for any rational $(a,b)$ except $(a,b)=(0,0)$

By Eisenstein's Irreducibility Criterion, $X^3 - 2$ is irreducible over the rationals. If there existed $a,b,c \in \mathbb{Q}$ , not all zero, such that $a + b\sqrt[3]{2} + c\sqrt[3]{4}= 0$ , then $f(X) = a + bX + cX^2 \in \mathbb{Q}(X)$ would be such that $f(\sqrt[3]{2}) = 0$ . Since $\sqrt[3]{2}$ is irrational, we must have $c \neq 0$ , and hence $f(X)$ is quadratic. Thus we must be able to write $X^3 - 2 = p(X)f(X) + q(X)$ for linear polynomials $p(X),q(X) \in \mathbb{Q}[X]$ . Since $p(\sqrt[3]{2}) =0$ and $\sqrt[3]{2}$ is irrational, we deduce that $p(X) \equiv 0$ , and hence that $f(X)$ divides $X^3 - 2$ . This is not possible. Thus if $a,b,c \in \mathbb{Q}$ are such that $a + b\sqrt[3]{2} + c\sqrt[3]{4}= 0$ , then $a=b=c=0$ .

Basically, if $\alpha = \sqrt[3]{2}$ , then the field $\mathbb{Q}(\alpha)$ obtained by adding $\alpha$ to $\mathbb{Q}$ is a $3$ -dimensional vector space over $\mathbb{Q}$ , and $1,\alpha,\alpha^2$ is a basis.