<s>[INST] Known that $a + b + c = 0$ and $a^3 + b^3 + c^3 = 27$. Find ABC! 

Known that
    $$a + b + c = 0$$
    $$a^3 + b^3 + c^3 = 27$$
    What is the value of $abc?$
A.) 1
B.) 0
C.) 7
D.) 8
E.) 10



My Work:
$$a + b = -c$$
$$a + c = -b$$
$$b + c = -a$$
Then
$$(a + b + c)(a + b + c)(a + b + c) = 0$$
Expanded into:
$$a^3 + b^3 + c^3 + 3a^2b+3b^2c+3a^2c+3ab^2 + 3bc^2+ 3ac^2+6abc = 0$$
Putting the Values:
$$27+3a^2(-a)+3b^2(-b)+3c^2(-c)+6abc = 0$$
$$27 -3(a^3 + b^3 + c^3) = -6abc$$
$$27-3(27) = -6abc$$
$$-54=-6abc$$
$$abc = 9$$
$9$ wasn't an option in the question. Am I missing something?
 [/INST] Your solution is correct.
$(a+b+c)^3=(a^3+b^3+c^3)+3a(b^2+c^2)+3b(a^2+c^2)+3c(a^2+b^2)+6abc$
$\implies 0=27+3ab^2+3ac^2+3ba^2+3bc^2+3ca^2+3cb^2+6abc$
$\implies 0 = 27+3b^2(a+c)+3a^2(b+c)+3c^2(a+b)+6abc$
$\implies 0 = 27+3b^2(-b)+3a^2(-a)+3c^2(-c)+6abc$
$\implies 0 = 27 - 3(a^3+b^3+c^3) +6abc$
$\implies 0 = 27 - 81 + 6abc$
$\implies 54 = 6abc$
$\implies abc = 9$
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