A) 0
B) \[-1\]
C) 1
D) \[-\left[ \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right]\]
Correct Answer: C
Solution :
| Here, the numerator is |
| \[={{a}^{2}}({{b}^{2}}-ac)\,\,({{c}^{2}}-ab)+{{b}^{2}}\,\,({{a}^{2}}-bc)({{c}^{2}}-ab)\]\[+{{c}^{2}}\,\,({{a}^{2}}-bc)\,\,({{b}^{2}}-ac)\] |
| \[={{a}^{2}}{{b}^{2}}{{c}^{2}}-{{a}^{3}}{{b}^{3}}-{{c}^{3}}{{a}^{3}}+c{{a}^{4}}b+{{b}^{2}}{{a}^{2}}{{c}^{2}}-{{a}^{3}}{{b}^{3}}\]\[-{{b}^{3}}{{c}^{3}}+{{a}^{4}}ca+{{b}^{2}}{{a}^{2}}{{c}^{2}}-{{c}^{3}}{{a}^{3}}-{{b}^{3}}{{c}^{3}}+b{{c}^{4}}a\] |
| \[=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-2{{a}^{3}}{{b}^{3}}-2{{a}^{3}}{{c}^{3}}-2{{b}^{3}}{{c}^{3}}+abc\,\,({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\]\[=3{{a}^{2}}{{b}^{2}}{{c}^{2}}-6{{a}^{2}}{{b}^{2}}{{c}^{2}}+abc\,\,({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\]\[=-3{{a}^{2}}{{b}^{2}}{{c}^{2}}+abc\,\,({{a}^{3}}+{{b}^{3}}+{{c}^{3}})\] |
| \[\because \] \[ab+bc+ac=0\] |
| \[\Rightarrow \]\[{{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{a}^{3}}{{c}^{3}}=3{{a}^{2}}{{b}^{2}}{{c}^{2}}\] |
| and denominator |
| \[=abc\,\,({{a}^{3}}+{{b}^{3}}+{{c}^{3}})-({{a}^{3}}{{b}^{3}}+{{a}^{3}}{{c}^{3}}+{{b}^{3}}{{c}^{3}})\] |
| \[=abc\,\,({{a}^{3}}+{{b}^{3}}+{{c}^{3}})-3{{a}^{2}}{{b}^{2}}{{c}^{2}}\](from above) |
| as Numerator = Denominator |
| \[\therefore \] Given function sum = 1 |
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