Question

Pinky says that zeroes of p(x)=x3-1 are 1,1,1 and sai teja say that one of the zeroes is 1 , remaining two zeroes are not real, to whom do you support give your reason

Answer 1

(x-1)(xv2+x+1)=0
X=1 or xv2+x+1=0
The discriminant is less than zero
Hence the roots are complex