Answer:
If (ca – b2) – (bc – a2) = (ab – c2) – (ca – b2)
bc – a2, ca – b2, ab – c2 are in A.P.
Consider LHS and RHS,
(ca – b2) – (bc – a2) = (ab – c2) – (ca – b2)
(a – b2 – bc + a2) = (ab – c2 – ca + b2)
(a – b) (a + b + c) = (b – c) (a + b + c)
a – b = b – c
a, b, c are in AP,
b – c = a – b
Hence, the given terms are in AP