Question 1132142
.


So, the question is:


<pre>
    Find the sum  {{{1/(root(3,1)+root(3,2)+root(3,4))}}} + {{{1/(root(3,4)+root(3,6)+root(3,9))}}} + {{{1/(root(3,9)+root(3,12)+root(3,16))}}}
</pre>

<U>Solution</U>


<pre>
Use the identity  {{{a^3-b^3}}} = {{{(a-b)*(a^2+ab + b^2)}}},   which gives you

                  {{{1/(a^2+ab + b^2)}}} = {{{(a-b)/(a^3-b^3)}}}.   <<<---=== it is how to "rationalize the fraction" in this case (!)


In this way,


{{{1/(root(3,1)+root(3,2)+root(3,4))}}} = {{{(root(3,2)-root(3,1))/(2-1)}}} = {{{root(3,2)-1}}}              <<<---=== in this case a= 2;  b= 1.


{{{1/(root(3,4)+root(3,6)+root(3,9))}}} = {{{(root(3,3)-root(3,2))/(3-2)}}} = {{{root(3,3)-root(3,2)}}};            <<<---=== in this case a= 3;  b= 2.


{{{1/(root(3,9)+root(3,12)+root(3,16))}}} = {{{(root(3,4)-root(3,3))/(4-3)}}} = {{{root(3,4)-root(3,3)}}}           <<<---=== in this case a= 4;  b= 3.


So, the long sum of three fractions is equal to


{{{root(3,2)-1}}} + {{{root(3,3)-root(3,2)}}} + {{{root(3,4)-root(3,3)}}} = {{{root(3,4)-1}}}.      <U>ANSWER</U>
</pre>

Solved.