Question 858284
THE FAST WAY:
If the product is {{{216}}} ,
{{{216=6^3=(2*3)^3=2^3*3^2}}}
The product of the three terms could be
{{{2*(2*3)*(2*3^2)=216}}} with the terms being 2, 6, and 18 (or 18, 6, and 2)
or
the product could be
{{{3*(3*2)*(3*2^2)=216}}} with the terms being 3, 6, and 12 (or 12, 6, and 3)
 
{{{18+2+6=26}}} while {{{12+6+3=21}}} , so the numbers are
{{{highlight(2)}}} , {{{highlight(6)}}} , and {{{highlight(18)}}} .
That gives as a solution quickly, without proving that it is the only solution.
 
THE EXPECTED WAY (with lots of formulas and calculations):
We will call the first term we are looking for {{{b[1]=b}}} ,
and the common ratio {{{r}}} .
The formula for term number {{{n}}} is {{{b[n]=b[1]*r^(n-1)}}} .
The second and third terms would be
{{{b[2]=b*r}}} and {{{b[3]=b*r^2}}} .
Their product would be {{{b*(b*r)*(b*r^2)=b^3*r*3=(br)^3}}}
We could calculate their sum as
{{{SUM=B+br+br^3=b(r^2+r+1)}}}
Otherwise, the formula for the sum of the first {{{n}}} terms is
{{{SUM[n]=b[1](r^n-1)/(r-1)}}} .
The sum of the three terms would be
{{{SUM=b(r^3-1)/(r-1)=(br^3-b)/(r-1)}}}
Our equations are
{{{b^3*r*3=216}}} or {{{(br)^3=216}}} and
{{{b(r^3-1)/(r-1)=26)}}} or {{{(br^3-b)/(r-1)=26)}}} or {{{b(r^2+r+1)=26}}} .
From {{{(br)^3=216}}} we get {{{br=root(3,210)}}}-->{{{br=6}}}-->{{{system(b=6/r,"or",r=6/b)}}}
If we substitute either one into {{{b(r^2+r+1)=26}}} we get an equation in one variable that we can solve.
{{{(6/r)(r^2+r+1)=26}}}-->{{{6(r^2+r+1)=26r}}}-->{{{6r^2+6r+6=26r}}}-->{{{6r^2+6r+6-26r=0}}}-->{{{6r^2-20r+6=0}}}<-->{{{3r^2-10r+3=0}}}
No matter how we solve {{{6r^2-20r+6=0}}} or {{{3r^2-10r+3=0}}}
we find {{{system(r=3,"or",r=1/3)}}} .
{{{system(r=3,b=6/r)}}}-->{{{system(r=3,b=6/3)}}}-->{{{highlight(system(r=3,b=2))}}}--> the terms , in order, are 2, 6, and 18.
{{{system(r=1/3,b=6/r)}}}-->{{{system(r=3,b=6/(1/3))}}}-->{{{system(r=1/3,b=6*3)}}}-->{{{highlight(system(r=1/3,b=18))}}}--> the terms , in order, are 18, 6, and 2.
The terms (regardles of order are {{{highlight(2)}}} , {{{highlight(6)}}} , and {{{highlight(18)}}} ,
and that is the only solution