Can anyone help me figure this problem out step by step? All help would be appreciated.
3/6-√2
√=radical
use the conjugate
so we have $\displaystyle 6-\sqrt{2}$ its conjugate is $\displaystyle 6+\sqrt{2}$
mult the numerator and denominator by the conjugate to get
$\displaystyle \frac{3}{6-\sqrt{2}} \cdot \frac{6+\sqrt{2}}{6+\sqrt{2}}=\frac{18+3\sqrt{2}}{ 6-2}=\frac{9}{2}+\frac{3}{4}\sqrt{2}$