What is the value of a^3 + b^3 + c^3 when a+b+c=0

This is an mcq in my notes, but I can't figure out the answer with any formula. If anyone knows the answer please tell, and specify how you solved the sum.

This was an MCQ, the options are:
0
abc
3abc
(ab+bc+ca)

Thanks in advance.

Using a=-b-c then substituting this for a^3

-b^3 -3b^2c-3bc^2 -c^3 +b^3 + c^3 = -3b^2c-3bc^2 = -3bc(b+c)

now b+c = -a so this gives -3bc(-1) = 3abc

In order to solve for 3 unknowns, you need 3 independent equations. So, the best you can do is to solve for 2 values in terms of the 3rd variable.

(a^3 + b^3 )+ c^3 = (a+b) (a^2-ab+b^2) +c^3

a+b+c=0
a+b= -c
-c( (a^2-ab+b^2) +c^3 = -a^2c +abc-cb^2 +c^3
= c^3-a^2c +abc-cb^2
= c(c^2-a^2) +bc(a-b)

= c(c-a) (c+a) + bc(a-b)
from
a+b+c=0

c+a= -b
= c(c-a) (-b) + bc(a-b)
= -bc^2+abc +abc-b^2c
= 2abc - bc ( c+b)

c+b= -a

= 2abc -bc(-a)
= 3abc

I plugged in random values to see what would happen.
A=2
B=3
C=-5

2^3 + 3^3 + (-5)^3 = 8 + 27 - 125 = -90 = 3abc

I've tried it with a bunch of values. I'm not sure why, but it works.

a + b + c = 0 ⇒ c = -(a + b)

a³ + b³ + c³ = a³ + b³ - (a + b)³
= a³ + b³ - a³ - 3a²b - 3ab² - b³
= -3ab(a + b)
= 3abc

Using statement 2 & the above formula:
Sufficient.
Ans: "B"

It's 0.

It should be self-explanatory.