Processing Math: Done

From Förberedande kurs i matematik 1

The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as (2+3)+6  and multiply the top and bottom of the fraction by the conjugate-like expression (2+3)−6 . Then, at least 6  will be squared away using the formula for the difference of two squares

1(2+3)+6(2+3)−6(2+3)−6=(2+3)−6(2+3)2−(6)2=(2+3)2−6(2+3)−6.

We expand the remaining quadratic, (2+3)2 , using the formula for the difference of two squares,

(2+3)2−6(2+3)−6=2+3−6(2)2+223+(3)2−6=2+3−62+223+3−6=26−12+3−6.

This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate 26+1 ,

26−12+3−6=26−12+3−626+126+1=(26)2−12(2+3−6)(26+1)=22(6)2−12226+21+326+31−626−61=46−122223+2+3223+3−2(6)2−6=24−12(2)23+2+2(3)22+3−26−6=23223+2+232+3−12−6=23(1+23)2+(22+1)3−12−6=2372+53−6−12.