Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is x^3-y^3 > x^2+xy+y^2?
1) x > y + 1
2) 0 < y < x
High-level question. Congrats, Max!
$${x^2} + xy + {y^2} = \underbrace {{x^2} + 2 \cdot x \cdot {y \over 2} + {{{y^2}} \over 4}}_{{{\left( {x + {y \over 2}} \right)}^{\,2}}} + {{3{y^2}} \over 4} \ge 0\,\,\,\,\,\,\,\left[ {\,\, \Rightarrow \,\,\,\,\,{x^2} + xy + {y^2} = 0\,\,\, \Leftrightarrow \,\,\left\{ \matrix{
\,{y^2} = 0 \hfill \cr
\,x + {y \over 2} = 0 \hfill \cr} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {x,y} \right) = \left( {0,0} \right)\,\,\,\,\,\,\left( * \right)\,\,} \right]$$
[The "
completing the squares" technique (above) is carefully explained in our course!]
$${x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\,\,\,\,\left( {**} \right)$$
$${x^3} - {y^3}\,\,\,\mathop > \limits^? \,\,\,{x^2} + xy + {y^2}\,$$
$$\left( 1 \right)\,\,x > y + 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{
\,\left( {x,y} \right) \ne \left( {0,0} \right)\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{x^2} + xy + {y^2} > 0\,\,\,\,\left( {***} \right) \hfill \cr
\,\,x - y > 1\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {***} \right)} \,\,\,\,\,\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\,\,\, > \,\,\,1 \cdot \left( {{x^2} + xy + {y^2}} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\,$$
$$\left( 2 \right)\,\,0 < y < x\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {3,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
The correct answer is therefore (A).
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
