If $$\sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1,$$        then what is the value of (1 - a²) (1 - b²) ?

Solution (By Examveda Team)

$$\eqalign{ & \sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1 \cr & {\text{Squaring both sides}} \cr & {a^2} + {b^2} + ab + {a^2} + {b^2} - ab + 2\sqrt {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} = 1 \cr & \sqrt {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} = \frac{1}{2} - \left( {{a^2} + {b^2}} \right) \cr & {\text{Again squaring both sides}} \cr & {a^4} + {b^4} + 2{a^2}{b^2} - {a^2}{b^2} = \frac{1}{4} + {\left( {{a^2} + {b^2}} \right)^2} - \left( {{a^2} + {b^2}} \right) \cr & {a^4} + {b^4} + {a^2}{b^2} = \frac{1}{4} + {a^4} + {b^4} + 2{a^2}{b^2} - {a^2} - {b^2} \cr & {a^2} + {b^2} - {a^2}{b^2} = \frac{1}{4}........\left( {\text{i}} \right) \cr & \Rightarrow \left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right) \cr & = 1 - {a^2} - {b^2} + {a^2}{b^2} \cr & = 1 - \left[ {{a^2} + {b^2} - {a^2}{b^2}} \right] \cr & = 1 - \frac{1}{4} \cr & = \frac{3}{4} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Let }}b = 0 \cr & \sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1 \cr & \sqrt {{a^2} + 0 + 0} + \sqrt {{a^2} + 0 + 0} = 1 \cr & a + a = 1 \cr & 2a = 1 \cr & a = \frac{1}{2} \cr & \left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right) \cr & = \left( {1 - \frac{1}{4}} \right)\left( {1 - 0} \right) \cr & = \frac{3}{4} \cr} $$