If $$A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}}$$     and $$B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}},$$     then what is the value of (A² + B²)²?

Solution (By Examveda Team)

$$\eqalign{ & A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}} \cr & = \frac{{{{\left( {0.6} \right)}^3} + {{\left( {0.2} \right)}^3}}}{{{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)}} \cr & = \frac{{\left( {0.6 + 0.2} \right)\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}}{{\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}} \cr & = 0.8 \cr & B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}} \cr & = \frac{{{{\left( {0.9} \right)}^3} - {{\left( {0.3} \right)}^3}}}{{{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)}} \cr & = \frac{{\left( {0.9 - 0.3} \right)\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}}{{\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}} \cr & = 0.6 \cr & \therefore \,{\left( {{A^2} + {B^2}} \right)^2} \cr & = {\left[ {{{\left( {0.8} \right)}^2} + {{\left( {0.6} \right)}^2}} \right]^2} \cr & = {\left( {0.64 + 0.36} \right)^2} \cr & = {\left( 1 \right)^2} \cr & = 1 \cr} $$