A) \[a+b+c=0\]
B) \[a+b+c=3abc\]
C) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=0\]
D) \[{{(a+b+c)}^{3}}=27abc\]
Correct Answer: D
Solution :
\[x+y+z=0\] \[\Rightarrow \] \[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz\] \[\therefore \] \[{{a}^{1/3}}+{{b}^{1/3}}+{{c}^{1/3}}=0\] \[\Rightarrow \] \[a+b+c=3{{a}^{1/3}}{{b}^{1/3}}{{c}^{1/3}}\] \[\Rightarrow \] \[a+b+c={{(3)}^{3}}abc\] \[\therefore \] \[{{(a+b+c)}^{3}}=27abc\]
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