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\frac{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{\left(\sqrt{6}-2\sqrt{3}\right)\left(\sqrt{6}+2\sqrt{3}\right)}
Rationalize the denominator of \frac{2\sqrt{3}}{\sqrt{6}-2\sqrt{3}} by multiplying numerator and denominator by \sqrt{6}+2\sqrt{3}.
\frac{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{\left(\sqrt{6}\right)^{2}-\left(-2\sqrt{3}\right)^{2}}
Consider \left(\sqrt{6}-2\sqrt{3}\right)\left(\sqrt{6}+2\sqrt{3}\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{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{6-\left(-2\sqrt{3}\right)^{2}}
The square of \sqrt{6} is 6.
\frac{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{6-\left(-2\right)^{2}\left(\sqrt{3}\right)^{2}}
Expand \left(-2\sqrt{3}\right)^{2}.
\frac{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{6-4\left(\sqrt{3}\right)^{2}}
Calculate -2 to the power of 2 and get 4.
\frac{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{6-4\times 3}
The square of \sqrt{3} is 3.
\frac{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{6-12}
Multiply 4 and 3 to get 12.
\frac{2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)}{-6}
Subtract 12 from 6 to get -6.
-\frac{1}{3}\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right)
Divide 2\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right) by -6 to get -\frac{1}{3}\sqrt{3}\left(\sqrt{6}+2\sqrt{3}\right).
-\frac{1}{3}\sqrt{3}\sqrt{6}-\frac{1}{3}\sqrt{3}\times 2\sqrt{3}
Use the distributive property to multiply -\frac{1}{3}\sqrt{3} by \sqrt{6}+2\sqrt{3}.
-\frac{1}{3}\sqrt{3}\sqrt{6}-\frac{1}{3}\times 3\times 2
Multiply \sqrt{3} and \sqrt{3} to get 3.
-\frac{1}{3}\sqrt{3}\sqrt{3}\sqrt{2}-\frac{1}{3}\times 3\times 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}.
-\frac{1}{3}\times 3\sqrt{2}-\frac{1}{3}\times 3\times 2
Multiply \sqrt{3} and \sqrt{3} to get 3.
-\sqrt{2}-\frac{1}{3}\times 3\times 2
Cancel out 3 and 3.
-\sqrt{2}-2
Cancel out 3 and 3.