Question 135224
{{{(8-sqrt(2))/(6-3sqrt(2))}}}


The trick is to multiply the fraction by 1 in the form of the conjugate of the denominator divided by itself.


If you have an expression of the form {{{a+b*sqrt(c)}}}, then its conjugate is {{{a-b*sqrt(c)}}}, and vice versa of course.


Here your denominator expression is {{{6-3sqrt(2)}}}, so the conjugate is {{{6+3sqrt(2)}}}, so you need to multiply your original fraction by 1 in the form of  {{{(6+3sqrt(2))/(6+3sqrt(2))}}}


{{{((8-sqrt(2))/(6-3sqrt(2)))((6+3sqrt(2))/(6+3sqrt(2)))}}}


Notice that multiplying a binomial times its conjugate is the reverse of factoring the difference of two squares -- so the result is the first term squared minus the second term squared.  Just use FOIL on the numerator expressions:


{{{(48+24sqrt(2)-6sqrt(2)-6)/(36-18)=(42+18sqrt(2))/18=(7+3sqrt(2))/3}}}


The result is a fraction with a rational number in the denominator, hence the name of the process.