# 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. – S.C.B. Jan 19 '17 at 14:03
• @S.C.B. Well, that's the end I guess... – Adola Jan 19 '17 at 14:04
• yes $$abc=9$$ is right – Dr. Sonnhard Graubner Jan 19 '17 at 19:47

## 3 Answers

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 $abc=9$.

Your solution is correct.

$(a+b+c)^3=(a^3+b^3+c^3)+3a(b^2+c^2)+3b(a^2+c^2)+3c(a^2+b^2)+6abc$

$\implies 0=27+3ab^2+3ac^2+3ba^2+3bc^2+3ca^2+3cb^2+6abc$

$\implies 0 = 27+3b^2(a+c)+3a^2(b+c)+3c^2(a+b)+6abc$

$\implies 0 = 27+3b^2(-b)+3a^2(-a)+3c^2(-c)+6abc$

$\implies 0 = 27 - 3(a^3+b^3+c^3) +6abc$

$\implies 0 = 27 - 81 + 6abc$

$\implies 54 = 6abc$

$\implies abc = 9$

if $a+b+c=0$

then ,

$a^3+b^3+c^3=3abc$ and $a^3+b^3+c^3=27$

equating both equations

$3abc=27$

therefore $abc=9$