Question

If x=√3-√2÷√3+√2 and y=√3+√2÷√3-√2, then show that x^2 + xy +y^2 = 99

Answer 1

thank u frnds ...hope this helps
@sumo

Answer 2

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ xy = 1 \\ \\ x + y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x + y = \frac{( { \sqrt{3} + \sqrt{2} })^{2} + ( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} \\ \\ x + y = 2(3 + 2) = 10 \\ \\ \\ {x}^{2} + {y}^{2} + xy = ( {x + y})^{2} - xy \\ = {10}^{2} - 1 \\ = 99$