Radical Rationalization?

We can rationalize the denominator..

$\Large \frac { \sqrt { 6 } \times \left( \sqrt { 2 } -\sqrt { 3 } \right) }{ \left( \sqrt { 2 } +\sqrt { 3 } \right) \times \left( \sqrt { 2 } -\sqrt { 3 } \right) } +\frac { 3\sqrt { 2 } \times \left( \sqrt { 6 } -\sqrt { 3 } \right) }{ \left( \sqrt { 6 } +\sqrt { 3 } \right) \times \left( \sqrt { 6 } -\sqrt { 3 } \right) } -\frac { 4\sqrt { 3 } \times \left( \sqrt { 6 } +\sqrt { 2 } \right) }{ \left( \sqrt { 6 } +\sqrt { 2 } \right) \times \left( \sqrt { 6 } -\sqrt { 2 } \right) }$

That is equal to

$\Large \frac { \sqrt { 6 } \times \left( \sqrt { 2 } -\sqrt { 3 } \right) }{ -1 } +\frac { 3\sqrt { 2 } \times \left( \sqrt { 6 } -\sqrt { 3 } \right) }{ 3 } -\frac { 4\sqrt { 3 } \times \left( \sqrt { 6 } +\sqrt { 2 } \right) }{ 4 }$

Again simplifying

$\large -\left[ \sqrt { 6 } \times \left( \sqrt { 2 } -\sqrt { 3 } \right) \right] +\left[ \sqrt { 2 } \times \left( \sqrt { 6 } -\sqrt { 3 } \right) \right] -\left[ \sqrt { 3 } \times \left( \sqrt { 6 } -\sqrt { 2 } \right) \right]$

$= \large -\sqrt { 12 } +\sqrt { 18 } +\sqrt { 12 } -\sqrt { 6 } -\sqrt { 18 } +\sqrt { 6 }$

All cancels off and the the answer will be $\boxed{0}$

rationalize denominator and u will find it will be reduced to $\boxed{0}$