A = $$\frac{{{x^8} - 1}}{{{x^4} + 1}}$$  and B = $$\frac{{{y^4} - 1}}{{{y^2} + 1}}.$$  If x = 2 and y = 9, then what is the value of A² + 2AB + AB²?

Solution(By Examveda Team)

$$\eqalign{ & A = \frac{{{x^8} - 1}}{{{x^4} + 1}} = \frac{{\left( {{x^4} + 1} \right)\left( {{x^4} - 1} \right)}}{{\left( {{x^4} + 1} \right)}} = {x^4} - 1 \cr & B = \frac{{{y^4} - 1}}{{{y^2} + 1}} = \frac{{\left( {{y^2} + 1} \right)\left( {{y^2} - 1} \right)}}{{\left( {{y^2} + 1} \right)}} = {y^2} - 1 \cr & x = 2,\,\,y = 9 \cr & A = {x^4} - 1 = {\left( 2 \right)^4} - 1 = 15 \cr & B = {y^2} - 1 = {\left( 9 \right)^2} - 1 = 80 \cr & {A^2} + 2AB + A{B^2} \cr & = {\left( {15} \right)^2} + 2 \times 15 \times 80 + 15 \times {\left( {80} \right)^2} \cr & = 15\left( {15 + 160 + 6400} \right) \cr & = 15 \times \left( {6575} \right) \cr & = 98625 \cr} $$