Lesson HOW TO rationalize a fraction by making its denominator free of cubic roots

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<H2>How to rationalize a fraction by making its denominator free of cubic roots</H2>

Sometimes you can see the fraction of the form like this &nbsp;&nbsp;{{{1/(root(3,2)-1)}}} &nbsp;&nbsp;containing the cubic root in the denominator. 
It is the natural desire to transform the fraction by making it free of the root in the denominator.
There is a standard way to get it.

The general recommendation is to multiply the denominator and the numerator of the fraction in the same number by choosing this number in such a way 
to make the denominator rational after applying the sum of cubes formula or the difference of cubes formula. 
To get information about the sum of cubes formula and the difference of cubes formula see the lessons 
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/The_sum_of_cubes_formula.lesson>The sum of cubes formula</A> and 
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/The_difference_of_cubes_formula.lesson>The difference of cubes formula</A> 
under the current topic in this site.

Below are examples that show how it works.

<H3>Example 1</H3>Transform the fraction &nbsp;&nbsp;{{{1/(root(3,2)-1)}}} &nbsp;&nbsp;to make it free of the cubic root in the denominator.

Solution

Let us multiply the denominator and the numerator of the fraction in the same number &nbsp;&nbsp;{{{(root(3,2))^2 + root(3,2) + 1}}}.
The value of the fraction will not change:

{{{1/(root(3,2)-1)}}} = {{{((root(3,2))^2 + root(3,2) + 1)/((root(3,2)-1)*((root(3,2))^2 + root(3,2) + 1))}}}.

Now apply the difference of cubes formula to the product &nbsp;&nbsp;{{{(root(3,2)-1)*((root(3,2))^2 + root(3,2) + 1)}}} &nbsp;&nbsp;in the denominator:

{{{(root(3,2)-1)*((root(3,2))^2 + root(3,2) + 1) = (root(3,2))^3 - 1 = 2 - 1 = 1}}}.

Finally, you have

{{{1/(root(3,2)-1)}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 + root(3,2) + 1)/((root(3,2)-1)*((root(3,2))^2 + root(3,2) + 1))}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 + root(3,2) + 1)/((root(3,2))^3 - 1) &nbsp;&nbsp;= &nbsp;&nbsp;((root(3,2))^2 + root(3,2) + 1)/(2 - 1)}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 + root(3,2) + 1)/1 &nbsp;&nbsp;= &nbsp;&nbsp;(root(3,2))^2 + root(3,2) + 1}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{root(3,4) + root(3,2) + 1}}}.

You got the same fraction (the same value, the same real number) without the cubic root in the denominator.
This is how this trick works.

The key to the success is in two elements:
&nbsp;&nbsp;&nbsp;&nbsp;1) the value of the fraction is not changing after multiplication the numerator and denominator of the fraction in the same number, and
&nbsp;&nbsp;&nbsp;&nbsp;2) try to choose the multiplicator in such a way to make the final expression of the denominator free of cubic roots.

Let us consider more examples.

<H3>Example 2</H3>Transform the fraction &nbsp;&nbsp;{{{1/(root(3,2)+1)}}} &nbsp;&nbsp;to make it free of the cubic root in the denominator.

Solution

Let us multiply the denominator and the numerator of the fraction in the same number &nbsp;&nbsp;{{{(root(3,2))^2 - root(3,2) + 1}}}.
The value of the fraction will not change:

{{{1/(root(3,2)+1)}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 - root(3,2) + 1)/((root(3,2)+1)*((root(3,2))^2 - root(3,2) + 1))}}}.

Now apply the sum of cubes formula to the product &nbsp;&nbsp;{{{(root(3,2)+1)*((root(3,2))^2 - root(3,2) + 1)}}} &nbsp;&nbsp;in the denominator:

{{{(root(3,2)+1)*((root(3,2))^2 - root(3,2) + 1) = (root(3,2))^3 + 1 = 2 + 1 = 3}}}.

Finally, you have

{{{1/(root(3,2)+1)}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 - root(3,2) + 1)/((root(3,2)+1)*((root(3,2))^2 - root(3,2) + 1))}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 - root(3,2) + 1)/((root(3,2))^3 + 1)}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 - root(3,2) + 1)/(2 + 1)}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{((root(3,2))^2 - root(3,2) + 1)/3}}} &nbsp;&nbsp;= &nbsp;&nbsp;{{{(root(3,4) - root(3,2) + 1)/3}}}.

You got the same fraction (the same value, the same real number) without the cubic root in the denominator.

<H3>Example 3</H3>Simplify the expression &nbsp;&nbsp;{{{1/(root(3,a^2) - root(3,a) + 1)}}}&nbsp;&nbsp; + &nbsp;&nbsp;{{{1/(root(3,a) + 1)}}}.

Solution

Let us start with simplifying the first term, which is the fraction &nbsp;&nbsp;{{{1/(root(3,a^2) - root(3,a) + 1)}}}.

I am going to multiply the denominator and the numerator of this fraction in &nbsp;&nbsp;{{{root(3,a)+1}}} &nbsp;&nbsp;and then to apply the sum of cubes formula.
After multiplication the fraction still remains equal to itself:

{{{1/(root(3,a^2) - root(3,a) + 1)}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{(root(3,a)+1)/((root(3,a^2) - root(3,a) + 1)*(root(3,a)+1))}}}.

Now apply the sum of cubes formula to the product in the denominator:

{{{(root(3,a^2) - root(3,a) + 1)*(root(3,a)+1)}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{(root(3,a))^3 + 1 = a+1}}}.

Thus you have the first term simplified

{{{1/(root(3,a^2) - root(3,a) + 1)}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{(root(3,a)+1)/((root(3,a^2) - root(3,a) + 1)*(root(3,a)+1)))}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{(root(3,a)+1)/(a+1)}}}.

Next, let us simplify the second term of the original expression, which is the fraction &nbsp;&nbsp;{{{1/(root(3,a) + 1)}}}.

This time multiply the denominator and the numerator of the fraction in &nbsp;&nbsp;{{{root(3,a^2) - root(3,a) + 1}}} &nbsp;&nbsp;and then apply the sum of cubes formula.
After multiplication the fraction still remains equal to itself:

{{{1/(root(3,a) + 1)}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{(root(3,a^2) - root(3,a) + 1)/((root(3,a) + 1)*(root(3,a^2) - root(3,a) + 1))}}}.

Now apply the sum of cubes formula to the product in the denominator:

{{{(root(3,a)+1)*(root(3,a^2) - root(3,a) + 1)}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{(root(3,a))^3 + 1 = a+1}}}.

Thus you have the second term simplified

{{{1/(root(3,a)+1)}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{((root(3,a^2) - root(3,a) + 1)/((root(3,a)+1)*(root(3,a^2) - root(3,a) + 1)))}}}&nbsp;&nbsp; = &nbsp;&nbsp;{{{(root(3,a^2) - root(3,a) + 1)/(a+1)}}}.

Actually, we transformed the first and the second terms of the original expression to the common denominator. 
The last step is to calculate the sum of the transformed first and the transformed second terms of the original expression. This sum is equal to

{{{(root(3,a)+1)/(a+1)}}} &nbsp;&nbsp;+&nbsp;&nbsp; {{{(root(3,a^2) - root(3,a) + 1)/(a+1)}}} &nbsp;&nbsp;=&nbsp;&nbsp; {{{(root(3,a^2) + 2)/(a+1)}}}

Thus the final simplified expression is &nbsp;&nbsp;{{{(root(3,a^2) + 2)/(a+1)}}}.

My other closely related lessons in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/HOW-TO-make-the-denominator-of-a-fraction-free-of-square-roots.lesson>HOW TO rationalize a fraction by making its denominator free of square roots</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Amazing-calculations-with-fractions-that-contain-quadratic-irrationalities-in-denominators.lesson>Amazing calculations with fractions that contain quadratic irrationalities in denominators</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/real-numbers/OVERVIEW-of-lessons-on-simplifying-and-rationalizing-functions-%28denominators%29.lesson>OVERVIEW of lessons on simplifying and rationalizing expressions and functions (denominators)</A>

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-I.