SOLUTION: Find the sum: {{{ 1/(sqrt(1)+sqrt(2)+sqrt(4))+1/(sqrt(4)+sqrt(6)+sqrt(9))+1/(sqrt(9)+sqrt(12)+sqrt(16))}}} P.S. to all tutors who are willing to solve this question, all

Algebra ->  Radicals -> SOLUTION: Find the sum: {{{ 1/(sqrt(1)+sqrt(2)+sqrt(4))+1/(sqrt(4)+sqrt(6)+sqrt(9))+1/(sqrt(9)+sqrt(12)+sqrt(16))}}} P.S. to all tutors who are willing to solve this question, all      Log On


   

Question 1132142: Find the sum:


P.S. to all tutors who are willing to solve this question, all the square root signs are supposed to be cube roots. I just did not know how to insert the equation. ALL THE SQUARE ROOT SIGNS ARE SUPPOSED TO BE CUBE ROOTS!!!!!!
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

So, the question is: 

    Find the sum  1%2F%28root%283%2C1%29%2Broot%283%2C2%29%2Broot%283%2C4%29%29 + 1%2F%28root%283%2C4%29%2Broot%283%2C6%29%2Broot%283%2C9%29%29 + 1%2F%28root%283%2C9%29%2Broot%283%2C12%29%2Broot%283%2C16%29%29

Solution 

Use the identity  a%5E3-b%5E3 = %28a-b%29%2A%28a%5E2%2Bab+%2B+b%5E2%29,   which gives you

                  1%2F%28a%5E2%2Bab+%2B+b%5E2%29 = %28a-b%29%2F%28a%5E3-b%5E3%29.   <<<---=== it is how to "rationalize the fraction" in this case (!)


In this way,


1%2F%28root%283%2C1%29%2Broot%283%2C2%29%2Broot%283%2C4%29%29 = %28root%283%2C2%29-root%283%2C1%29%29%2F%282-1%29 = root%283%2C2%29-1              <<<---=== in this case a= 2;  b= 1.


1%2F%28root%283%2C4%29%2Broot%283%2C6%29%2Broot%283%2C9%29%29 = %28root%283%2C3%29-root%283%2C2%29%29%2F%283-2%29 = root%283%2C3%29-root%283%2C2%29;            <<<---=== in this case a= 3;  b= 2.


1%2F%28root%283%2C9%29%2Broot%283%2C12%29%2Broot%283%2C16%29%29 = %28root%283%2C4%29-root%283%2C3%29%29%2F%284-3%29 = root%283%2C4%29-root%283%2C3%29           <<<---=== in this case a= 4;  b= 3.


So, the long sum of three fractions is equal to


root%283%2C2%29-1 + root%283%2C3%29-root%283%2C2%29 + root%283%2C4%29-root%283%2C3%29 = root%283%2C4%29-1.      ANSWER

Solved.