As we know, (a + b + c)3 = a3 + b3 + c3 + 3(a + b)(b + c)(a + c) Then, here, a = x - 1, b = y - 2 and c = z - 3 Now, according to the question : a3 + b3 + c3 + 3(a + b)(b + c)(a + c) = 0 So, (a + b + c)3 = 0 (a + b + c) = 0 Then, (x - 1 + y - 2 + z - 3) = 0 Therefore, (x + y + z) = 6 So, the correct option is (C) : 6.