Rationalize the denominator of $\frac{1+\sqrt{3}}{1-\sqrt{3}}$. When you write your answer in the form $A+B\sqrt{C}$, where $A$, $B$, and $C$ are integers, what is $ABC$?
I notice that the denominator of the fraction is 1 minus the square root of 3, which is an irrational number. To rationalize the denominator, I want to multiply the fraction by something that will make the denominator a rational number. I recall that if I multiply a binomial by its conjugate, I get the difference of two squares. The conjugate of 1 minus the square root of 3 is 1 plus the square root of 3, which is the same as the numerator of the fraction. So I multiply the fraction by $\frac{1+\sqrt{3}}{1+\sqrt{3}}$, which is equivalent to multiplying by 1, so the value of the fraction does not change. In the numerator, I use the distributive property to get $(1+\sqrt{3})(1+\sqrt{3})=1+2\sqrt{3}+3=4+2\sqrt{3}$. In the denominator, I use the difference of two squares formula to get $(1-\sqrt{3})(1+\sqrt{3})=1-3=-2$. So the fraction becomes $\frac{4+2\sqrt{3}}{-2}$. To simplify the fraction, I divide both the numerator and the denominator by -2, and I get $\frac{-2-\sqrt{3}}{1}$, which is the same as $-2-\sqrt{3}$. Now I have the answer in the form $A+B\sqrt{C}$, where $A=-2$, $B=-1$, and $C=3$. To find $ABC$, I multiply these three numbers and I get $-2\times -1\times 3=6$. 