<s>[INST] 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?
 [/INST] 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.
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