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\frac{\sqrt{3}\left(6+\sqrt{3}\right)}{\left(6-\sqrt{3}\right)\left(6+\sqrt{3}\right)}
Rationalize the denominator of \frac{\sqrt{3}}{6-\sqrt{3}} by multiplying numerator and denominator by 6+\sqrt{3}.
\frac{\sqrt{3}\left(6+\sqrt{3}\right)}{6^{2}-\left(\sqrt{3}\right)^{2}}
Consider \left(6-\sqrt{3}\right)\left(6+\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{\sqrt{3}\left(6+\sqrt{3}\right)}{36-3}
Square 6. Square \sqrt{3}.
\frac{\sqrt{3}\left(6+\sqrt{3}\right)}{33}
Subtract 3 from 36 to get 33.
\frac{6\sqrt{3}+\left(\sqrt{3}\right)^{2}}{33}
Use the distributive property to multiply \sqrt{3} by 6+\sqrt{3}.
\frac{6\sqrt{3}+3}{33}
The square of \sqrt{3} is 3.