not crackable

Is it 1?

a²a²+(a)(a)+bc + b²b²+(b)(b)+ca + c²c²+(c)(c)+ab

= a²a²-a(b+c)+bc + b²b²-b(c+a)+ca + c²c²-c(a+b)+ab

= a²(a-b)(a-c) + b²(b-c)(b-a) + c²(c-a)(c-b)

= - a²(b-c)(a-b)(b-c)(c-a) - b²(c-a)(a-b)(b-c)(c-a) - c²(a-b)(a-b)(b-c)(c-a)

= - a²b - a²c + b²c - b²a + c²a - c²b(a-b)(b-c)(c-a)

= - a²b + a²c - b²c + b²a - c²a + c²b- a²b + a²c - b²c + b²a - c²a + c²b

= 1

the second last step of bipin sir's can be simplified
by putting a=b
the expresion reduces to 0
hence a-b is a factor
since expression is symmetric and cyclic the same holds true for b-c and c-a
hence answer is 1[1]

xyz...the condition is given a≠b≠c...kaise assume karen phir?