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?

No, the question is wrong, and your answer is right. However, it might be easier by recalling the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
From $a+b+c=0$, we have $abc=9$. 
