Let’s simplify:
i. 
ii. 
iii. 
iv. 
v. 
vi. 
i. (a – b) + (b – c) + (c – a)
= a – b + b – c + c – a
= a – a + b – b + c – c
= 0
ii. (a + b)(a – b) + (b – c)(b + c) + (c + a)(c –a)
We know, (a + b)(a – b) = a2 – b2, therefore
= a2 – b2 + b2 – c2 + c2 – a2
= 0
iii. 
Cancelling out common terms from numerator and denominator,
= (x2 – y2)(x2 + y2)
= x4 – y4 [∵ a2 – b2 = (a – b)(a + b), Here a = x2, b = y2]
iv. a(b – c) + b(c – a) + c(a – b)
= ab – ac + bc – ab + ac – bc
= ab – ab + bc – bc + ac – ac
= 0
v. x2(y2 – z2) + y2(z2 – x2) + z2(x2 – y2)
= x2y2 – x2z2 + y2z2 – x2y2 + x2z2 – y2z2
= x2y2 – x2y2 + y2z2 – y2z2 + x2z2 – x2z2
= 0
vi. (x3 + y3)(x3 – y3) + (y3 + z3)(y3 – z3) + (z3 – x3)(z3 + x3)
We know, (a + b)(a – b) = a2 – b2, therefore
= (x3)2 –(y3)2 + (y3)2 – (z3)2 + (z3)2 – (x3)2
= x6 – y6 + y6 – z6 + z6 – x6
= 0
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