Given condition, (a+bi)3=a−bi i.e. a3+i3b3+3a2bi−3ab2i2=a−bi ⇒a3−ib3+i3a2b−3ab2=a−bi (a3−3ab2)−i(b3−3a2b)=a−bi Equating the coefficient, we obtain a3−3ab2=a and b3−3a2b=b ⇒a(a2−3b2)=a and b(b2−3a2)=b ⇒a2−3b2=1 .....(1) and b2−3a2=1.....(2) Use Eq. (i) in (ii) b2−3(1+3b2)=1 ⇒b2−3−9b2=1or8b2+4=0 This gives b2=−