If ∣∣ ∣∣abcbcacab∣∣ ∣∣=k(a+b+c)(a2+b2+c2−bc−ca−ab), then k=
If a + b + c = 0, then prove the following
(a) (b + c) (b − c) + a(a + 2b) = 0
(b) a(a2 − bc) + b(b2 − ca) + c(c2 − ab) = 0
(c) a(b2 + c2) + b(c2 + a2) + c(a2 + b2) = −3abc
(d)