Question:
If
\[ a^{1/3}+b^{1/3}+c^{1/3}=0, \] then
(a) \[ a+b+c=0 \]
(b) \[ (a+b+c)^3=27abc \]
(c) \[ a+b+c=3abc \]
(d) \[ a^3+b^3+c^3=0 \]
Solution:
Let
\[ x=a^{1/3}, \quad y=b^{1/3}, \quad z=c^{1/3} \]
Then
\[ x+y+z=0 \]
Using identity:
\[ x^3+y^3+z^3=3xyz \quad \text{when} \quad x+y+z=0 \]
Substituting back:
\[ a+b+c = 3\left(a^{1/3}b^{1/3}c^{1/3}\right) \]
\[ a+b+c = 3(abc)^{1/3} \]
Now cubing both sides:
\[ (a+b+c)^3 = 27abc \]
Hence, the correct answer is:
\[ \boxed{(a+b+c)^3=27abc} \]