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).6036
B).6039
C).6042
D).6048

Answer 1

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.