Question 413641
{{{(1 - (sqrt(3))/3)/(1 + (sqrt(3))/3)}}}
First let's get rid of the "little" fractions within the "big" fraction. This can be done by multiplying the numerator and denominator of the "big" fraction by the lowest common denominator (LCD) of the "little" denominators. Since both "little" denominators are 3's the LCD is 3:
{{{((1 - (sqrt(3))/3)/(1 + (sqrt(3))/3))(3/3)}}}
To multiply we will use the Distributive Property in both the numerator and denominator:
{{{(3 - sqrt(3))/(3 + sqrt(3))}}}
Next we need to rationalize the denominator. Since the denominator has two terms this is a little tricky. What we will be doing is taking advantage of the {{{(a+b)(a-b) = a^2-b^2}}} pattern.
This pattern shows us how to take a two-term expression, (a+b) or (a-b), and multiply it by something an get nothing but perfect squares, {{{a^2-b^2}}}
Our denominator has a "+' between the terms so it will play the role of (a+b), with "a" being 3 and "b" being {{{sqrt(3)}}}. To rationalize it we will multiply the numerator and denominator by (a-b) or {{{(3-sqrt(3))}}}:
{{{((3 - sqrt(3))/(3 + sqrt(3)))((3-sqrt(3))/(3-sqrt(3)))}}}
In the denominator we get the {{{a^2-b^2}}} as planned. In the numerator we have, in effect (a-b)(a-b). So we can use the {{{(a-b)(a-b) = a^2-2ab+b^2}}} pattern to multiply it quickly:
{{{((3)^2 -2(3)(sqrt(3))+(sqrt(3))^2)/((3)^2 - (sqrt(3))^2)}}}
which simplifies as follows:
{{{(9 -6sqrt(3)+3)/(9 - 3)}}}}
{{{(12 -6sqrt(3))/6}}}
This fraction can be reduced:
{{{(6(2 -sqrt(3)))/6}}}
{{{(cross(6)(2 -sqrt(3)))/cross(6)}}}
leaving:
{{{2-sqrt(3)}}}