Cubic Equation

#Problem 6  
The cubic equation x^3+2x-1 = 0 has exactly one real root r. Note that 0.4 < r < 0.5.

(a)Prove that an increasing sequence of positive integers a1,a2,a3,.....exists, such that 1/2=r^a1+r^a2+r^a3+.... is possible

(b) Prove that the sequence that you found in part (a) is the unique increasing sequence
with the above property.

Solution Scheme and approach
(a)
As r is a root then we may write
r^3+2r-1=0 =>1-r^3=2r=>r/(1-r^3)=1/2=>r*(1-r^3)^-1=1/2
By expanding
r*(1+r^3+r^6+r^9+....)=1/2
As |r|<1 the series converges
=>r+r^4+r^7+r^10+....=1/2 equating with r^a1+r^a2+r^a3+..=1/2
we can write {a1,a2,a3....}::{1,4,7....}

(b)Proof by contradiction
If possible let another for another series b1,b2,b3....               r^b1+r^b2+r^b3+...=r^a1+r^a2+r^a3+...=1/2                                                                                                       canceling equal terms
r^x1+r^x2+r^x3+...=r^y1+r^y2+r^y3+...=N [N>0]............(1)
without losing the generality we can consider y1>x1 =>r^y1>r^x1
r^x1<=r^x1+r^x2+r^x3+..=r^y1+r^y2+r^y3+....
Dividing both sides by r^x1
1<=r^p1+r^p2+r^p3+......[yi-xi=pi]...........(2)
As 0<1/2 =>1<1/(1-r)<2 and p1>=1 then,                                                                                                  From (2)
1<=r^p1+r^p2+r^p3+...<=r+r^2+r^3+....=r/1-r=1/(1-r)-1<1
A contradiction.
Hence only one sequence exists.