<s>[INST] If $a+b+c=0$, then $a^3+b^3+c^3$ is ... $0$? $1$? $a^3b^3c^3$? $3abc$? Many mistakes in this post. ( See comments below). I let it as it is, as an example of what shouldn't be done.


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

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?

 [/INST] 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 $2$
</s>