?
$$\frac{{\frac{1}{3}.\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4}.\frac{1}{4} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{5}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4} + \frac{1}{5}.\frac{1}{5} - \left( {\frac{1}{3}.\frac{1}{4} + \frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{3}} \right)}}{\text{ is?}}$$
Answer & Solution
Correct Answer:
Option
C
$$\frac{{\frac{1}{3}.\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4}.\frac{1}{4} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{5}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4} + \frac{1}{5}.\frac{1}{5} - \left( {\frac{1}{3}.\frac{1}{4} + \frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{3}} \right)}}$$
A3 + B3 + C3 - 3ABC = (A + B + C)(A2 + B2 + C2 - AB - BC - CA)
$$\therefore \frac{{{{\left( {\frac{1}{3}} \right)}^3} + {{\left( {\frac{1}{4}} \right)}^3} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + {{\left( {\frac{1}{5}} \right)}^3}}}{{{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}}}$$
$$ = \frac{{\left( {\frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right)\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}{{\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}$$
$$\eqalign{ & = \frac{{20 + 15 + 12}}{{60}} \cr & = \frac{{47}}{{60}} \cr} $$
A3 + B3 + C3 - 3ABC = (A + B + C)(A2 + B2 + C2 - AB - BC - CA)
$$\therefore \frac{{{{\left( {\frac{1}{3}} \right)}^3} + {{\left( {\frac{1}{4}} \right)}^3} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + {{\left( {\frac{1}{5}} \right)}^3}}}{{{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}}}$$
$$ = \frac{{\left( {\frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right)\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}{{\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}$$
$$\eqalign{ & = \frac{{20 + 15 + 12}}{{60}} \cr & = \frac{{47}}{{60}} \cr} $$
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