A) \[abc\]
B) \[\frac{1}{abc}\]
C) \[{{a}^{2}}{{b}^{2}}{{c}^{2}}\]
D) \[\frac{1}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}\]
Correct Answer: B
Solution :
Let the value of \[a=3,b=2\]and \[c=1\] \[\frac{1}{a(a-b)\,(a-c)}+\frac{1}{b(b-a)(b-c)}+\frac{1}{c(c-a)\,(c-b)}\] \[=\frac{1}{3\times 1\times 2}+\frac{1}{2\times (-1)\times 1}+\frac{1}{1\times (-2)\times (-1)}=\frac{1}{6}\] Now put the value of \[a=3,\,\,b=2,\,\,c=1\] in Option \[=\frac{1}{abc}=\frac{1}{3\times 2\times 1}=\frac{1}{6}\] \[\therefore \,\,\,\,\,\frac{1}{a(a-b)\,(a-c)}+\frac{1}{b(b-a)\,(b-c)}+\] \[\frac{1}{c(c-a)\,(c-b)}=\underline{\frac{\mathbf{1}}{\mathbf{abc}}}\]
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