Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$ Given that $\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}$, where $a,b,c$ are integers. Show $a=b=c=0$
Do I use modular arithmetic?
 A: Start with $b \sqrt[3]{2} + c \sqrt[3]{4} = -a$. Cube this relation to find another
equation of the form $B \sqrt[3]{2} + C \sqrt[3]{4} = -A$ for rationals A, B, C. Eliminating
the cube root of 4 from these equations will tell you that the cube root of 2 is rational.
This contradiction shows that $a = b = c = 0$.
A: Hint $\ $ If so then $\rm\:x = \sqrt[3]{2}\:$ would be a root of $\rm\:f = a+b\:x+c\:x^2\:$ and $\rm\: g = x^3-2\:$ so also a root of their gcd $\rm = e f + h g\:$ (by Bezout), contra $\rm\:x^3-2\:$ is irreducible over $\rm\:\mathbb Q\:$ by the rational root test.
Alternatively, if $\rm\:w = \sqrt[3]{2}\:$ is a nonrational root of a quadratic then there exists a conjugation automorphism $\rm\:x\mapsto x'\:$ on $\rm\mathbb Q(w)\:$ with fixed field $\rm \mathbb Q,\:$ so taking the norm $\rm\:xx'$ of $\rm\:w^3 = 2\:$ yields $\rm\:(ww')^3 = 4\:$ for $\rm\:ww'\in \mathbb Q,\:$ contradiction.
A: By standard results in field theory, given the element $\alpha$ a root of the polynomial $x^3 - 2$, $\alpha$ by definition is algebraic over $\Bbb{Q}$ so that $[\Bbb{Q}(\alpha):\Bbb{Q}]$ is finite (in particular equal to the degree of $x^3 - 2$ that is three). Viewing $\Bbb{Q}(\alpha)$ as a vector space of dimension three over $\Bbb{Q}$, there is the usual basis for $\Bbb{Q}(\alpha)$ given by $1, \alpha, \alpha^2$. It is not hard to prove that this is a basis using the division algorithm and the fact that the kernel of the evaluation map $\Bbb{Q}[x] \longrightarrow \Bbb{Q}(\alpha)$ is the principal ideal $(x^3 - 2)$. Taking $\alpha = \sqrt[3]{2}$ shows that the numbers $1, \sqrt[3]{2}, (\sqrt[3]{2})^2$ are linearly independent over $\Bbb{Q}$, so that in particular the only solution in integers to the equation
$$a + b\sqrt[3]{2} + c(\sqrt[3]{2})^2= 0$$
is the trivial solution.
A: A decade late but here's a nice elementary proof:
Take $x = a$, $y = \sqrt[3]{2}b$ and $z = \sqrt[3]{4}c$
Note the identity
$$x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)((x - y)^2 + (y - z)^2 + (x - z)^2)$$
If $x = y = z$, we have $a = b = c = 0$.
Otherwise $x + y + z = 0 \iff x^3 + y^3 + z^3 - 3xyz = 0$
Substitute back to obtain
$$\iff a^3 + 2b^3 + 4c^3 - 6abc = 0$$
Consider a solution $(a_0, b_0, c_0) \not = (0, 0, 0)$, take modulo $2$ and note $a_0 = 2a_1$ for some integer $a_1$.
Then,
$$8a_1^3 + 2b_0^3 + 4c_0^3 - 12a_1b_0c_0 = 0$$
Divide by 2 to obtain
$$b_0^3 + 2c_0^3 + 4a_1^3 - 6b_0c_0a_1 = 0$$
So $(b_0, c_0, a_1)$ is also a solution! We can keep applying this transformation 2 more times to get $(a_1, b_1, c_1)$ is also a solution, where similarly $(b_1, c_1) = (2b_0, 2c_0)$. However this means that we can divide factors of 2 out of our original solution an arbitrary amount of times, contradiction.
A: Let $\alpha=\sqrt[3]{2}$
Then $m_{\alpha}({x})=x^3-2$ is the minimal polynomial of $\alpha$ over $\Bbb{Q}$.
$f(x)=a+bx+cx^2$ is an anhilating polynomial of $\sqrt[3]{2}$ as $f(\alpha) =0$
Hence $m_{\alpha}(x) \mid f(x) $
$x^3-2\mid a+bx+cx^2$ implies $a=b=c=0$
