Given that $a, b, c \in \mathbb{Z}$, then $$8a^{3}+8b^{3}+8c^{3}-12a^{2}+6a-24abc+12bc \neq 1$$ I tried to factor this somehow, but I don't think it's factorable. After writing it as $8a^{3}+8b^{3}+8c^{3}=1+(2a-1)(12bc+6a)$, I noticed that $a=\frac{1}{2}$, $b=-c$ verifies this equation. If we can prove that these are the only solutions, we are done.
After this observation, I tried dividing the polynomial in $a$ by $(2a-1)$, which yields $$(2a-1)(4a^2-4a-12bc+1)=8b^3+8c^3$$ I thought I was onto something when i subtracted $(2a-1)(4a^2-4a-12bc+1)=8b^3+8c^3$ from $8a^{3}+8b^{3}+8c^{3}=1+(2a-1)(12bc+6a)$, but that just gives me $(2a-1)(4a^2+2a+1)=8a^3-1$, which is just a true statement.
I would appreciate if asomeone could give me some insight in how to solve this.
Hint:
All of the coefficients in $$8a^{3}+8b^{3}+8c^{3}-12a^{2}+6a-24abc+12bc$$ are even numbers.