If $a+b+c=0$, then $a^3+b^3+c^3$ is ... $0$? $1$? $a^3b^3c^3$? $3abc$?

Question

Many mistakes in this post. ( See comments below). I let it as it is, as an example of what shouldn't be done.

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If $a+b+c=0$, then $a^3+b^3+c^3 = \ldots $

A. $\;0\quad$ B. $\;1\quad$ C. $\;a^3b^3c^3\quad$ D. $\;3abc$

Source: 4/12/2020, Competitive Exams Reasoning Sample Paper 3- Translation in Hindi, Kannada, Malayalam, Marathi, Punjabi, Sindhi, Sindhi, Tamil, Telgu - Examrace. Downloaded from examrace.com

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I can only see the pitfall consisting in inferring that all 3 numbers must be equal to 0.

What I can conclude from the premise is that one of the 3 numbers is the additive inverse of the sum of the 2 others.

Admitting it is number $c$, we get

$$a+b+c = 0= (a+b) + \left( - (a+b) \right) \tag{1}$$

In that case $$c^3 = [- (a+b)]^3 = - (a+b) (a+b)(a+b) = - ( a^3 +2a^2b+2ab^2+b^3) \tag{2}$$

So $$\begin{align} a^3+b^3+c^3 &= a^3+b^3 - ( a^3 +2a^2b+2ab^2+b^3) \tag{3} \\ &= a^3+b^3 - a^3 - 2a^2b- 2ab^2- b^3 \tag{4}\\ &= 2a^2b - 2ab^2 \tag{5} \\ &=2 ( a^2b - b^2a) \tag{6} \\ &= 2 ( a) (ab-b^2) \tag{7} \\ &= 2 ( a) (b) (a-b) \tag{8} \\ &= 2 ( a) (b) (- c) \quad\text{[ Since $c = -(a+b) = b - a = - (a-b) $]} \tag{9} \\ &= - 2 ( a) (b) (c) \tag{10} \end{align}$$

However, this isn't one of the possible answers.

What did I miss? Was I wrong in supposing that I could take any number $a$, $b$, or $c$ to play the role of additive inverse of the sum of the two others?

Answer 1

Let $a=1$, $b=1$, $c=-2$. So $a+b+c=0$ and $a^3+b^3+c^3=-6$.

Therefore, choices (1), (2), and (3) (whose values are $0$, $1$, and $-8$, respectively) are wrong. But choice (4) has $3abc=-6$. Therefore choice (4) is the correct answer.

Answer 2

Normally $$a^3+b^3+c^3 = (a+b+c)^3-3(a+b)(a+c)(b+c)$$ If $a+b+c = 0$

Then $a^3+b^3+c^3 = -3(a+b)(a+c)(b+c)$ Setting $c = -a-b$

$$a^3+b^3+c^3 = -3ab(b+a)$$ $$a^3+b^3+c^3 = 3abc$$

Recheck you expansion of $(a+b)^3$ in the second line of your work, you've replace $3$ with $2$

Answer 3

I believe you did the expansion of $(a+b)(a+b)(a+b)$ wrong. It should be $a^3 + 3ab^2 + 3a^2b + b^3$.