Answer :

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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