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Thursday, June 20, 2013

Algebraic Operations - Radicals, 5

Category: Algebra

"Published in Newark, California, USA"

Perform the indicated operations


Solution:

Consider the given equation above


The first thing that we have to do is to examine the radicals first if they can simplify or not. As a rule in Mathematics, all radicals must be simplified as much as we can.

At the first term, 16 is not a perfect cube. The factors of 16 are 8 and 2. 8 is a perfect cube.

At the second term, the denominator contains a radical. We need to eliminate the radical sign at the denominator by rationalization of the denominator. Multiply both the numerator and denominator by x2 so that the denominator becomes a perfect cube which is x3

At the third term, the denominator contains a radical also. We need to eliminate the radical sign at the denominator by rationalization of the denominator. Multiply both the numerator and denominator by 2 so that the denominator becomes a perfect cube which is 8. 

Hence, the given equation above becomes




Take the cube root of the numbers inside the radicals that are perfect cube, we have 



Since all the terms inside the radicals are the same, then we can combine them and therefore, the final answer is