Given equation is, x3−3x2+3x−9=0 ⇒(x−3)(x2+3)=0 So one of root is 3 Now, x2+3=0 taking x=1+2w, we have (1+2ω)2+3 =1+4ω2+4ω+3 =1+(4ω2+4ω+4)−1 =1+4(ω2+ω+1)−1 =0; or ω2+ω+1=0 Similarly, 1+2ω2 also satisfies given equation So roots are; 3,1+2ω and 1+2ω2