$\displaystyle \frac{8}{2 - \sqrt{2}} $
is this right, $\displaystyle \frac{8}{2 - \sqrt{2}} X \frac{2 + \sqrt{2}}{2 + \sqrt{2}} $ = 16+$\displaystyle 8\sqrt{2} $, as the numerator and 6 as the dinominator?
$\displaystyle \frac{8}{2 - \sqrt{2}} $
is this right, $\displaystyle \frac{8}{2 - \sqrt{2}} X \frac{2 + \sqrt{2}}{2 + \sqrt{2}} $ = 16+$\displaystyle 8\sqrt{2} $, as the numerator and 6 as the dinominator?
$\displaystyle \frac{8}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{8(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} $
$\displaystyle = \frac{(16 + 8\sqrt{2})}{(2^2 + 2\sqrt{2} - 2\sqrt{2} + \sqrt{2}^2)} = \frac{(16 + 8\sqrt{2})}{(4-2)} = \frac{(16 + 8\sqrt{2})}{2} = 8 + 4\sqrt{2}$