What is the greatest value... The sum of the first three terms of a geometric progression is 21, and the sum of their squares is 1281. What is the greatest value can be the sum of the cubes?
 A: Let $a,ar,ar^2$ be the three terms
So, $a+ar+ar^2=21$
So, $(a)^2+(ar)^2+(ar^2)^2=1281\implies a(r^2+r+1)a(r^2-r+1)=1281$
On division, $a-ar+ar^2=\frac{1281}{21}=61$
So, $ar=-20,a+ar^2=41$
$a^3+(ar)^3+(ar^2)^3=a^3+(ar^2)^3+(-20)^3$
$=(a+ar^2)^3-3a\cdot ar^2(a+ar^2)+(-20)^3$
$=(41)^3-3(-20)^2(41)+(-20)^3$
A: Hints:


*

*You have two distinct equalities and two unknowns so there are a finite number of solutions satisfying the equalities.  So start by solving the equalities  

*If $S_1$ is the sum of the three terms and $S_2$ is the sum of their squares you could start by looking at $\dfrac{S_1^2-S_2}{2S_1}$ to see if this gives you anything useful 

*$21^2 = 441 \lt 1281$, so not all the terms in the geometric progression can be positive
A: Call the terms $\dfrac{b}{r}$, $b$, and $br$.  Square the sum. We get 
$$\frac{b^2}{r^2}+b^2+b^2r^2 +2b\left(\frac{b}{r}+b+br\right).$$
Thus $441= 1281+2b(21)$. It follows that $b=-20$. The rest is routine.
If we want to do the rest prettily, use
$$\frac{1}{r^3}+r^3=\left(\frac{1}{r}+r\right)^3-3\left(\frac{1}{r^2}+r^2\right).$$
