Освободиться от иррациональности в знаменателе дроба: 3+√2+√3 / 3-√2-√3

[latex]= \frac{(3+ \sqrt{2}+ \sqrt{3} )(3- \sqrt{2} + \sqrt{3} )}{(3- \sqrt{2} - \sqrt{3} )(3- \sqrt{2} + \sqrt{3} )} = \frac{(3+ \sqrt{3}+ \sqrt{2})(3+ \sqrt{3}- \sqrt{2} ) }{(3- \sqrt{2} )^2 -( \sqrt{3} )^2}= \\ \\ = \frac{(3+ \sqrt{3} )^2-( \sqrt{2} )^2}{9-6 \sqrt{2}+2-3 }= \frac{9+6 \sqrt{3}+3-2 }{8-6 \sqrt{2} }= \frac{10+6 \sqrt{3} }{8-6 \sqrt{2} }= \\ \\ = \frac{2(5+3 \sqrt{3} )}{2(4-3 \sqrt{2} )}= \frac{5+3 \sqrt{3} }{4-3 \sqrt{2} }= \\ \\ = \frac{(5+3 \sqrt{3} )(4+3 \sqrt{2} )}{(4-[latex] \sqrt{2}(4+3 \sqrt{2} )}= \frac{20+12 \sqrt{3}+15 \sqrt{2}+9 \sqrt{6} }{4^2-(3 \sqrt{2} )^2}= \\ \\ = \frac{20+12 \sqrt{3}+15 \sqrt{2}+9 \sqrt{6} }{16-18}= -10 -6 \sqrt{3} -7.5 \sqrt{2}-4.5 \sqrt{6} [/latex])()} [/latex]