Question 793959
Since {{{(4sqrt(2)-sqrt(6))(4sqrt(2)+sqrt(6))=(4sqrt(2))^2-(sqrt(6))^2=4^2*2-6=16*2-6=32-6=26}}},
to get rid of the square roots in the denominator, we can multiply numerator and denominator times {{{(4sqrt(2)+sqrt(6))}}}:
{{{(3sqrt(2)+sqrt(6))/(4sqrt(2)-sqrt(6))}}} = {{{(3sqrt(2)+sqrt(6))(4sqrt(2)+sqrt(6))/(4sqrt(2)-sqrt(6))/(4sqrt(2)+sqrt(6))}}} = {{{(3sqrt(2)*4sqrt(2)+3sqrt(2)*sqrt(6)+sqrt(6)*4sqrt(2)+sqrt(6)*sqrt(6))/26}}}=


{{{(3*4*2+3sqrt(2)*sqrt(6)+sqrt(6)*4sqrt(2)+6)/26=(24+7sqrt(2)*sqrt(6)+6)/26=(30+7sqrt(2*6))/26=(30+7sqrt(4*3))/26=(30+7sqrt(4)*sqrt(3))/26=(30+7*2sqrt(3))/26=(30+14sqrt(3))/26}}}