Problem: The expression 1β11+3+11β31-\dfrac{1}{1+\sqrt{3}}+\dfrac{1}{1-\sqrt{3}}1β1+3β1β+1β3β1β equals
Answer Choices:
A. 1β31-\sqrt{3}1β3β
B. 111
C. β3-\sqrt{3}β3β
D. 3\sqrt{3}3β
E. 1+31+\sqrt{3}1+3β Solution:
1β (1+3)(1β3)β1β (1β3)+1β (1+3)(1+3)(1β3)=1β3\dfrac{1 \cdot(1+\sqrt{3})(1-\sqrt{3})-1 \cdot(1-\sqrt{3})+1 \cdot(1+\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=1-\sqrt{3}(1+3β)(1β3β)1β (1+3β)(1β3β)β1β (1β3β)+1β (1+3β)β=1β3β.