Question

# For the given example choose the correct alternative and fill in the blanks:\eqalign{ & {\left( {2012} \right)^3} + {\left( {2013} \right)^3} + {\left( {2014} \right)^3} - 3 \times 2012 \times 2013 \times 2014 \cr & = \left( {......} \right) \times \left\{ {{{\left( {2012} \right)}^2} + {{\left( {2013} \right)}^2} + {{\left( {2014} \right)}^2} - 2012 \times 2013 - 2013 \times 2014 - 2014 \times 2012} \right\} \cr}A).6036B).6039C).6042D).6048

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Hint: We are going to solve this problem by using formula of ${a^3} + {b^3} + {c^3}$
We have ${a^3} + {b^3} + {c^3} = (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ac) + 3abc$
$\Rightarrow {a^3} + {b^3} + {c^3} - 3abc = (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ac)$
Let a=2012, b=2013, and c=2014, Taking L.H.S from the given equation
$\Rightarrow {(2012)^3} + {(2013)^3} + {(2014)^3} - 3 \times 2012 \times 2013 \times 2014$
It is in the form of ${a^3} + {b^3} + {c^3} - 3abc$
$= \left( {2012 + 2013 + 2014} \right)\left( {{{(2012)}^2} + {{(2013)}^2} + {{(2014)}^2} - 2012 \times 2013 - 2013 \times 2014 - 2014 \times 2012} \right)$$= (6039)\left( {{{(2012)}^2} + {{(2013)}^2} + {{(2014)}^2} - 2012 \times 2013 - 2013 \times 2014 - 2014 \times 2012} \right)$
$\therefore$6039 is the number required in the given blank.

Note:
Here we solved the given problem using basic algebraic formula ${a^3} + {b^3} + {c^3} = (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ac) + 3abc$
We compared the given problem with this formula and simplified the expression to get the required value.