Question

x1/3 + y1/3 + z1/3 = 0, then (A) x + y + z = 3 xyz (B) x + y + z = 0 (C) ( x + y + z)3 = 27 xyz (D) x3 + y3 + z3 = 0 (A) x + y + z = 3 xyz (B) x + y + z = 0 (C) ( x + y + z)3 = 27 xyz (D) x3 + y3 + z3 = 0

Answer 1

Option (B) is correct
As x+y+z =0