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\frac{4\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}
Rationalize the denominator of \frac{4}{3-\sqrt{3}} by multiplying numerator and denominator by 3+\sqrt{3}.
\frac{4\left(3+\sqrt{3}\right)}{3^{2}-\left(\sqrt{3}\right)^{2}}
Consider \left(3-\sqrt{3}\right)\left(3+\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{4\left(3+\sqrt{3}\right)}{9-3}
Square 3. Square \sqrt{3}.
\frac{4\left(3+\sqrt{3}\right)}{6}
Subtract 3 from 9 to get 6.
\frac{2}{3}\left(3+\sqrt{3}\right)
Divide 4\left(3+\sqrt{3}\right) by 6 to get \frac{2}{3}\left(3+\sqrt{3}\right).
\frac{2}{3}\times 3+\frac{2}{3}\sqrt{3}
Use the distributive property to multiply \frac{2}{3} by 3+\sqrt{3}.
2+\frac{2}{3}\sqrt{3}
Cancel out 3 and 3.