Answer
$\dfrac{3-\sqrt{3}}{6}=\dfrac{1}{3+\sqrt{3}}$
Work Step by Step
$\dfrac{3-\sqrt{3}}{6}$
Multiply the numerator and the denominator of the given expression by $3+\sqrt{3}$, which is the conjugate of the numerator:
$\dfrac{3-\sqrt{3}}{6}\cdot\dfrac{3+\sqrt{3}}{3+\sqrt{3}}=...$
Evaluate the products:
$...=\dfrac{3^{2}-(\sqrt{3})^{2}}{6(3+\sqrt{3})}=\dfrac{9-3}{6(3+\sqrt{3})}=\dfrac{6}{6(3+\sqrt{3})}=\dfrac{1}{3+\sqrt{3}}$
