If x,y,z are distinct,then  1+x3    x2    11+y3    y2    11+z3    z2    1=0 ,when xyz is

Since each element of C1 is the sum of two elements, putting the determinant as sum of two determinants, we getΔ=x3x2xy3y2yz3z2z+1x2x1y2y1z2z=xyzx2x1y2y1z2z1+1x2x1y2y1z2z=−(xyz+1)1xx21yy21zz2=−(xyz+1)(x−y)(y−z)(z−x)(x+y+z)Since Δ=0, x, y, z all are distinct, wo have xyz + 1 = 0or xyz = -1.