How do you rationalize the denominator and simplify #3/(sqrt[6] – sqrt[3])#?

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Answer 1 · 2016-03-17T21:52:55

#sqrt(6)+sqrt(3)#

Known: #a^2-b^2 = (a-b)(a+b)#

Using this principle:

Multiply by 1 but in the form of #1=(sqrt(6)+sqrt(3))/(sqrt(6)+sqrt(3))#

#(3(sqrt(6)+sqrt(3)))/((sqrt(6)-sqrt(3))times(sqrt(6)+sqrt(3))#

#=(3sqrt(6)+3sqrt(3))/((sqrt(6))^2-(sqrt(3))^2)#

#=(cancel(3)sqrt(6)+cancel(3)sqrt(3))/cancel(3)#

#=sqrt(6)+sqrt(3)#