we have `(7sqrt3)/(sqrt10+sqrt3)-(2sqrt5)/(sqrt6+sqrt5)-(3sqrt2)/(sqrt15+3sqrt2)`
now`(7 sqrt3)/(sqrt10+sqrt3)xx(sqrt10-sqrt3)/(sqrt10-sqrt3)=(7sqrt30-21)/((sqrt10)^(2)-(sqrt3)^(2))`
[ using identity , `(a-b)(a+b) = a^(2)-b^(2)]`
`(7sqrt30-21)/(10-3)=(7sqrt30-3)/7=sqrt30-3`
` (2sqrt5)/(sqrt6+sqrt5)=(2sqrt5)/((sqrt5+sqrt5))xx((sqrt6-sqrt5))/((sqrt6-sqrt5))`
[by rationslisation ]
`(2sqrt30-10)/((sqrt6)^(2)-(sqrt5)^(2)) " " ["using identity", (a-b)(a+b)=a^(2)-b^(2)]`
`(2sqrt30-10)/(6-5)=2sqrt30-10`
` (3sqrt2)/(sqrt15+3sqrt2)=(3sqrt2)/(sqrt15+3sqrt2)xx(sqrt15-3sqrt2)/(sqrt15-3sqrt2) " " ["by rationalisation"]`
`(3sqrt30-18)/((sqrt15)^(2)-(3sqrt2)^(2)) " " ["using identity",(a-b) (a+b)=a^(2)-b^(2)]`
`(3(sqrt30-6))/(15-18)=(3(sqrt30-6))/(-3)=(-sqrt30+6)`
`6-sqrt30`
from Eq. (i) `(7sqrt3)/(sqrt10+sqrt3)-(2sqrt5)/(sqrt6+sqrt5)-(3sqrt2)/(sqrt15+3sqrt2)`
`= (sqrt30-3)-(2sqrt30-10)-(6-sqrt30)`
`sqrt30-3-2sqrt30+10-6+sqrt30`
`2sqrt30-2sqrt30+10-9=1`
