Question 1209387
<br>
The given series is<br>
a+ar+ar^2+ar^3+...<br>
The sum of the series is 3:<br>
[1] {{{a/(1-r)=3}}}<br>
The series consisting of the cubes of the terms of the given series is<br>
a^3r^3+a^3r^6+a^3r^9+...<br>
The sum of that series is 5:<br>
[2] {{{a^3/(1-r^3)=5}}}<br>
To find the common ratio r, solve [1] for a in terms of r and substitute in [2].<br>
{{{a=3(1-r)}}}<br>
{{{(3(1-r))^3/(1-r^3)=5}}}<br>
{{{27(1-3r+3r^2-r^3)=5(1-r^3)}}}<br>
{{{27-81r+81r^2-27r^3=5-5r^3}}}<br>
{{{22r^3-81r^2+81r-22=0}}}<br>
By inspection, r=1 is one solution to that equation.  However r=1 produces an infinite geometric series that has no sum.<br>
Use synthetic division to remove the root x=1 to find the other two roots.<br><pre>

  1 | 22 -81  81 -22
    |     22 -59  22
    +---------------
      22 -59  22   0</pre>
The remaining quadratic is {{{22x^2-59x+22}}}<br>
Use the quadratic formula to find that the other two solutions are<br>
{{{(59+-sqrt(1545))/44}}}<br>
There are two possible values for the common ratio r:<br>
ANSWERS:
{{{(1/44)(59+sqrt(1545))}}}
{{{(1/44)(59-sqrt(1545))}}}<br>