Answer :
Given that
[tex] \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2}} = a + b\sqrt{6} [/tex]
Rationalising the denominator by its conjugate
[tex] \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2}} * \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}+\sqrt{2}} = a + b\sqrt{6} [/tex]
[tex]\frac{ (\sqrt{3}+\sqrt{2})^2 }{(\sqrt{3})^2-(\sqrt{2})^2}= a + b\sqrt{6} [/tex]
[tex]\frac{3+2+2\sqrt{3}.\sqrt{2}}{3-2}=a+b\sqrt{6}[/tex]
[tex]5 + 2\sqrt{6} = a + b\sqrt{6}[/tex]
Therefore, a = 5 and b = 2
[tex] \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2}} = a + b\sqrt{6} [/tex]
Rationalising the denominator by its conjugate
[tex] \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}-\sqrt{2}} * \frac{ \sqrt{3}+\sqrt{2} }{\sqrt{3}+\sqrt{2}} = a + b\sqrt{6} [/tex]
[tex]\frac{ (\sqrt{3}+\sqrt{2})^2 }{(\sqrt{3})^2-(\sqrt{2})^2}= a + b\sqrt{6} [/tex]
[tex]\frac{3+2+2\sqrt{3}.\sqrt{2}}{3-2}=a+b\sqrt{6}[/tex]
[tex]5 + 2\sqrt{6} = a + b\sqrt{6}[/tex]
Therefore, a = 5 and b = 2