Let ω≠1, be a cube root of unity, and a,b∈R.Statement-1: a3+b3=(a+b)aω+bω2aω2+bωStatement-2: x3−1=(x−1)xω2−ωxω−ω2 for each x∈R.

We have –xxω2−ωxω−ω2    =x2ω3−xω2−xω4+ω3    =x2−ω2+ωx+1     =x2+x+1 ∵ω2+ω=−1∴ (x−1)xω2−ωxω−ω2                         =(x−1)x2+x+1=x3−1Thus, Statement-2 is true. Replacing x by –x, we get(−x)3−1=(−x−1)−xω2−ω−xω−ω2⇒      x3+1=(x+1)xω2+ωxω+ω2Taking conjugate of both the sides, we get        x3+1=(x+1)xω+ω2xω2+ω                                                           [∴  ω=ω2]If b=0, , statement-1 is clearly true. Suppose b≠0.Replacing x by a/b we get      ab3+1=ab+1abw+w2abw2+w⇒ a3+b3=(a+b)aw+bw2aw2+bwThus, statement-1 is also true and statement-2 is a correct explanation for it.