A) $ -1 $
B) $ abc $
C) $ 1 $
D) $ 0 $
Correct Answer: D
$ \Rightarrow $ $ a^{3}+b^{3}+c^{3}-3abc=0 $
$ \Rightarrow $ $ a^{3}+b^{3}+c^{3}=3abc $ (ii) Now, $ \frac{1}{a^{3}}+\frac{1}{{{(b+c)}^{3}}}=\frac{{{(b+c)}^{3}}+a^{3}}{a^{3}{{(b+c)}^{3}}} $ $ =\frac{b^{3}+c^{3}+3b^{2}c+3bc^{2}+a^{3}}{a^{3}{{(b+c)}^{3}}} $ $ =\frac{a^{3}+b^{3}+c^{3}+3bc^{2}+3bc^{2}}{a^{3}{{(b+c)}^{3}}} $ $ =\frac{3abc+3b^{2}c+3bc^{2}}{a^{3}{{(b+c)}^{3}}} $ [from Eq. (ii)] $ =\frac{3bc(a+b+c)}{a^{3}{{(b+c)}^{3}}}=0 $ [from Eq. (i)] Similarly, $ [ \frac{1}{b^{3}}+\frac{1}{{{(a+c)}^{3}}} ]=0 $ and $ [ \frac{1}{c^{3}}+\frac{1}{{{(a+b)}^{3}}} ]=0 $ Now, $ \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{{{(a+b)}^{3}}}+\frac{1}{{{(b+c)}^{3}}}+\frac{1}{{{(c+a)}^{3}}}=0 $ Alternate Method $ a+b+c=0 $
$ \Rightarrow $ $ a+b=-c $
$ \therefore $ $ \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{{{(a+b)}^{3}}}+\frac{1}{{{(b+c)}^{3}}}+\frac{1}{{{(c+a)}^{3}}} $ $ =\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}-\frac{1}{c^{3}}-\frac{1}{a^{3}}-\frac{1}{b^{3}}=0 $
BETA
