Question

(3√2÷√3+√6)-(4√3÷√6+√2)+(√6÷√2+√3)
with following steps

Answer 1

(3√2 /√3 + √6) - (4√3/√6+√2) + (√6 / √2+√3)
rationalize the denominator and then you will get -
=(3√6 - 6√3/3 - 6) - (12√2 - 4√6/ 6-2) + (2√3 - 3√2/ 2-3)
=( -3√6 + 6√3/3) - (12√2 - 4√6/ 4) + (-2√3 + 3√2/ 1)
we will take lcm now of the denominator
={4( -3√6 + 6√3) - (12√2 - 4√6) + (-2√3 + 3√2)}/12
= {-12√6 +24√3 - 36√2 + 12√6 - 24√3 + 36√2}/12
= 0/12
= 0