<s>[INST] If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$, is $a=b=c=0$? If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$ , then is it true that $a=b=c=0$ ? I was thinking of infinite descent but can't actually proceed , please help. Thanks in advance 
 [/INST] We may assume that $a$, $b$, and $c$ are relatively prime.  If not, we can divide out by the cube of the common factor.  Observe that $$c^3=2(3abc-2a^3-b^3),$$ so $c$ is even.  Let $c=2n$.
Then, the equation becomes $4a^3+2b^3+8n^3=12abn$.  Therefore,
$$
2b^3=4(3abn-a^3-2n^3).
$$
Hence, $b$ is even.  Let $b=2m$.
Then, the equation becomes $4a^3+16m^3+8n^3=24amn$.  Therefore,
$$
4a^3=8(3amn-2m^3-n^3).
$$
Hence $a$ is even, a contradiction.
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