Does $a^3 + 2b^3 + 4c^3 = 6abc$ have solutions in $\mathbb{Q}$ Does $a^3 + 2b^3 + 4c^3 = 6abc$ have solutions in $\mathbb{Q}$?
This is not a homework problem. Indeed, I have no prior experience in number theory and would like to see a showcase of common techniques used to solve problems such as this. Thanks
Edit Apart from $a=b=c=0$.
 A: HINT Make use of
$$x^3 + y^3 + z^3 -3xyz = \left(x+y+z \right) \left(x^2+y^2+z^2-xy-yz-zx \right)$$where $x=a$, $y=b\sqrt[3]{2}$ and $z=c \sqrt[3]{4}$.
A: First, note that  $(a,b,c)$ is a solution if and only if  $(ka,kb,kc)$ is; hence we may assume $a,b,c$ are integers, with no common factor (divide by that common factor if necessary).
Because $6abc, 2b^3+4c^3$ are even, so is $a^3$ and hence $a$.  Write $a=2a'$ and we have $$8(a')^3+2b^3+4c^3=12a'bc$$ and hence $$4(a')^3+b^3+2c^3=6a'bc$$
By similar logic, $b$ is even, so write $b=2b'$ and we have $$4(a')^3+8(b')^3+2c^3=12a'b'c$$
But now $$2(a')^2+4(b')^3+c^3=6a'b'c$$ and hence $c$ is even.  Hence $a,b,c$ are all even; this contradicts $a,b,c$ having no common factor.

Followup: The same proof works if the coefficients $\{1,2,4,6\}$ are replaced by $\{\alpha_1, \alpha_2, \alpha_3,\alpha_4\}$ so long as there is some prime $p$ with $\nu_p(\alpha_1)=0, \nu_p(\alpha_2)=1, \nu_p(\alpha_3)=2, \nu_p(\alpha_4)\ge 1$. (Here $\nu_p(\cdot)$ denotes the p-adic valuation). For example, apart from $(0,0,0)$, there are no rational solutions to $$7a^3+15b^3+18c^3=45abc$$ where here $p=3$. 

$~$

Double followup: The same proof works with $n$ variables $$\alpha_0a_0^n+\alpha_1a_1^n+\cdots+\alpha_{n-1}a_{n-1}^n=\alpha_n(a_0a_1\cdots a_{n-1})$$
  provided that $\nu_p(\alpha_i)=i$ (for $0\le i\le n-1$) and $\nu_p(\alpha_n)\ge 1$.

A: As it happens, the cubic form $C(a,b,c)=a^3+2b^3+4c^3-6abc$ is the norm form for the extension $K=\mathbb Q(\root 3 \of 2)$ over $\mathbb Q$. That is, if you look at a general element of $K$, say $a + b\root3\of2+c\root3\of2^2$, and take its field-theoretic norm, what you get is exactly $C(a,b,c)$. Now, the norm doesn’t vanish on an algebraic extension, except at zero. So, just because $C$ happens to be a norm form, you can say immediately that the trivial zero is the only one.
How did I spot this? By having done lots of examples, first by hand over many years, then, more recently, with symbolic algebra programs.
A: if negative numbers are accepted, (57,63,-156) is OK
A: (Just to give to user $71815$ a distinct way to solve).
We use $(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc$.
Taking in account that $$\left(\frac{a}{\sqrt 6}\right)^3+\left(\frac{b}{\sqrt3}\right)^3+\left(\frac{c\sqrt2}{\sqrt3}\right)^3=abc$$ we get after calculation and equating to zero the coefficients of $\sqrt[3]2$ and $\sqrt[3]4$ the equations $$c(a^2-2ac-2b^2c)=0$$  $$abc=0$$
