Question

$If a = (3-√5)/(3+√5) and b = = (3+√5)/(3-√5) ,then find a2 - b2.$

Answer 1

$a=\frac{3-\sqrt5}{3+\sqrt5};\ b=\frac{3+\sqrt5}{3-\sqrt5}\\\\a^2-b^2=\left(\frac{3-\sqrt5}{3+\sqrt5}\right)^2-\left(\frac{3+\sqrt5}{3-\sqrt5}\right)^2=(*)\\\\\left(\frac{3-\sqrt5}{3+\sqrt5}\right)^2=\frac{9-6\sqrt5+5}{9+6\sqrt5+5}=\frac{14-6\sqrt5}{14+6\sqrt5}\\\\\left(\frac{3+\sqrt5}{3-\sqrt5}\right)^2=\frac{9+6\sqrt5+5}{9-6\sqrt5+5}=\frac{14+6\sqrt5}{14-6\sqrt5}$

$(*)=\frac{14-6\sqrt5}{14+6\sqrt5}-\frac{14+6\sqrt5}{14-6\sqrt5}=(*)=\frac{(14-6\sqrt5)^2}{(14+6\sqrt5)(14-6\sqrt5)}-\frac{(14+6\sqrt5)^2}{(14+6\sqrt5)(14-6\sqrt5)}\\\\=\frac{(14-6\sqrt5)^2-(14+6\sqrt5)^2}{14^2-(6\sqrt5)^2}=\frac{(14-6\sqrt5-14-6\sqrt5)(14-6\sqrt5+14+6\sqrt5)}{196-180}\\\\=\frac{(-12\sqrt5)(28)}{16}=-3\sqrt5\cdot7=\boxed{-21\sqrt5}$