Given that the roots of the equation $x^3-9x^2+bx-216=0$ are consecutive terms in a geometric sequence, find the value of b. Given that the roots of the equation $x^3-9x^2+bx-216=0$ are consecutive terms in a geometric sequence, find the value of b.
First of all I found that 6 is a root of the equation: $\frac{α}{r}+α+αr=α^3,   α^3=216$ (product of the roots)
Then I used the remainder theorem and I got $b=54$. The answer should be $b=18$, so instead I factorized the equation using standard division and still got the same result. 
Any hints how to get the correct solution?
 A: Let $\alpha$, $\beta$ and $\gamma$ be our roots.
Thus, $$\beta^2=\alpha\gamma,$$
$$\alpha+\beta+\gamma=9$$ and $$\alpha\beta\gamma=216.$$
Thus, from the first and the third we obtain:
$$\beta^3=216$$ or
$$\beta=6.$$
Thus, $$a+\gamma=3$$ and
$$\alpha\gamma=36.$$
Id est,
$$b=\alpha\beta+\alpha\gamma+\beta\gamma=6(\alpha+\gamma)+\alpha\gamma=6\cdot3+36=54,$$
which says that you are right and your book is wrong.
It's interesting that for $b=54$ we have
$$(x-6)(x^2-3x+36)=0,$$
which has no three real roots, but these roots are  geometric sequence.
Thus, or in the given should be $b$ is a real number or we need to work also with other two cases:
$$\beta=6(\cos120^{\circ}+i\sin120^{\circ});$$
$$\beta=6(\cos240^{\circ}+i\sin240^{\circ}).$$
A: If the roots are consecutive terms in a geometric sequence, we have
$\alpha_1 = a$, $\alpha_2 = ar$, $\alpha_3 = ar^2$, 
giving $a^3r^3 = -216$
Inspection tells us that there are various possibilities for $a$ and $r$, involving permutations of $-2$ and $-3$ (given that $216 = 2^33^3$).
Expanding out should give the desired result.
