<s>[INST] Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$. 
Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.

My Attempt
$$a^3+b^3+c^3-3abc=2017$$
$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$
Now, $a+b+c$ can't be equal to $1$ as $a, b, c$ are natural numbers.
So, $$a+b+c=2017$$ $$a^2+b^2+c^2-ab-bc-ca=1$$
How should I proceed after this?
 [/INST] 
It was an easy question, just one important result and a bit of number theory was needed
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