| If \[a+b+c\]then find the value of \[\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}+\frac{1}{{{(a+b)}^{3}}}+\frac{1}{{{(b+c)}^{3}}}+\frac{1}{{{(c+a)}^{3}}}.\] [SSC (CGL) 2014] |
A) \[-1\]
B) \[abc\]
C) \[1\]
D) \[0\]
Correct Answer: D
Solution :
| Given, \[a+b+c=0\] ... (i) |
| We know that, \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)\]\[({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+bc+ca)\] |
| \[\Rightarrow \]\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\] |
| \[\Rightarrow \] \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\] (ii) |
| Now,\[\frac{1}{{{a}^{3}}}+\frac{1}{{{(b+c)}^{3}}}=\frac{{{(b+c)}^{3}}+{{a}^{3}}}{{{a}^{3}}{{(b+c)}^{3}}}\] |
| \[=\frac{{{b}^{3}}+{{c}^{3}}+3{{b}^{2}}c+3b{{c}^{2}}+{{a}^{3}}}{{{a}^{3}}{{(b+c)}^{3}}}\] |
| \[=\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3b{{c}^{2}}+3b{{c}^{2}}}{{{a}^{3}}{{(b+c)}^{3}}}\] |
| \[=\frac{3abc+3{{b}^{2}}c+3b{{c}^{2}}}{{{a}^{3}}{{(b+c)}^{3}}}\] [from Eq. (ii)] |
| \[=\frac{3bc\,\,(a+b+c)}{{{a}^{3}}{{(b+c)}^{3}}}=0\] [from Eq. (i)] |
| Similarly, \[\left[ \frac{1}{{{b}^{3}}}+\frac{1}{{{(a+c)}^{3}}} \right]=0\]and \[\left[ \frac{1}{{{c}^{3}}}+\frac{1}{{{(a+b)}^{3}}} \right]=0\] |
| 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 |
| \[a+b+c=0\] |
| \[\Rightarrow \] \[a+b=-\,\,c\] |
| \[\therefore \]\[\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}+\frac{1}{{{(a+b)}^{3}}}+\frac{1}{{{(b+c)}^{3}}}+\frac{1}{{{(c+a)}^{3}}}\] |
| \[=\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}-\frac{1}{{{c}^{3}}}-\frac{1}{{{a}^{3}}}-\frac{1}{{{b}^{3}}}=0\] |
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