Question 1205016
<pre>Thought I'd do it by the sum formula, to see if it's harder or easier.
</pre>
The sum of the first three terms of a geometric sequence is 8...<pre> 
{{{S[n]}}}{{{""=""}}}{{{(a[1](r^n-1))/(r-1)}}}

{{{S[3]}}}{{{""=""}}}{{{(a[1](r^3-1))/(r-1)}}}

{{{8}}}{{{""=""}}}{{{(a[1](r^3-1))/(r-1)}}}</pre>
and the sum of the first six terms is 12.<pre>
{{{S[6]}}}{{{""=""}}}{{{(a[1](r^6-1))/(r-1)}}}

{{{12}}}{{{""=""}}}{{{(a[1](r^6-1))/(r-1)}}}

{{{system(8=(a[1](r^3-1))/(r-1),12=(a[1](r^6-1))/(r-1))}}}

{{{system(8(r-1)=a[1](r^3-1),
12(r-1)=a[1](r^6-1))}}}

Divide the 2nd equation by the first equation:

{{{12/8=(r^6-1)/(r^3-1)}}}

{{{3/2=((r^3-1)(r^3+1))/(r^3-1)}}}

{{{3/2=(cross((r^3-1))(r^3+1))/cross(r^3-1)}}}

{{{3/2=r^3+1}}}

{{{3=2r^3+2}}}

{{{1=2r^3}}}

{{{1/2=r^3}}}

{{{root(3,1/2)=r}}}

{{{root(3,expr(1/2)*expr(2/2)*expr(2/2))=r}}}

{{{root(3,4)/2=r}}}

Edwin</pre>