Answer
$\displaystyle \frac{4}{8-\sqrt{3}}=\frac{32+4\sqrt{3}}{61}$
Work Step by Step
$\displaystyle \frac{4}{8-\sqrt{3}}$
$\qquad$ ...rationalize the denominator by multyplying both the numerator and the denominator
with the conjugate of $8-\sqrt{3}$, which is $8+\sqrt{3}$.
$=\displaystyle \frac{4}{8-\sqrt{3}}\cdot\frac{8+\sqrt{3}}{8+\sqrt{3}}\qquad$ ...apply $(a+b)(a-b)=a^{2}-b^{2}$ in the denominator ($a=8,\ b=\sqrt{3}$).
$=\displaystyle \frac{32+4\sqrt{3}}{8^{2}-(\sqrt{3})^{2}}\qquad$ ...simplify ($8^{2}=64,\ (\sqrt{3})^{2}=3$).
$=\displaystyle \frac{32+4\sqrt{3}}{64-3}\qquad$ ...add like terms.
$=\displaystyle \frac{32+4\sqrt{3}}{61}$
