If the equation `x^3 + ax^2+ bx+216=0` has three roots in GP then b/a is equal to

Explanation

Let `alpha/r,alpha,alphar` be the roots. Then
`alpha^(3)=-216`
Again `alpha^(2)/r+alpha^(2)r+alpha^(2)=b`
`alpha^(2)(1+r+1/r)=b` (2)
and `alpha(1+r+1/r)=-a` (3)
On dividing (2) by (3), we get
`alpha=-b/a`
or `alpha^(3)=-b^(3)/a^(3)` (4)
From (1) and (4), `(b/a)^(3)=216`
or `b/a=-6`