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\frac{3\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}
Rationalize the denominator of \frac{3\sqrt{6}}{\sqrt{3}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{2}.
\frac{3\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}
Consider \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{3\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{3-2}
Square \sqrt{3}. Square \sqrt{2}.
\frac{3\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{1}
Subtract 2 from 3 to get 1.
3\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)
Anything divided by one gives itself.
3\sqrt{6}\sqrt{3}-3\sqrt{6}\sqrt{2}
Use the distributive property to multiply 3\sqrt{6} by \sqrt{3}-\sqrt{2}.
3\sqrt{3}\sqrt{2}\sqrt{3}-3\sqrt{6}\sqrt{2}
Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.
3\times 3\sqrt{2}-3\sqrt{6}\sqrt{2}
Multiply \sqrt{3} and \sqrt{3} to get 3.
9\sqrt{2}-3\sqrt{6}\sqrt{2}
Multiply 3 and 3 to get 9.
9\sqrt{2}-3\sqrt{2}\sqrt{3}\sqrt{2}
Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.
9\sqrt{2}-3\times 2\sqrt{3}
Multiply \sqrt{2} and \sqrt{2} to get 2.
9\sqrt{2}-6\sqrt{3}
Multiply -3 and 2 to get -6.