The value of the determination ∣∣ ∣ ∣∣b2−abb−cbc−acab−a2a−bb2−abbc−acc−aab−a2∣∣ ∣ ∣∣=
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)