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Value of (a – c) [(a – b)2 + (b – c)2 – (a – b) (b – c)] + (c – a)3 is equal to
(a – c) [(a – b)2 + (b – c)2 – (a – b) (b – c)] + (c – a)3Take (a – c) [(a – b)2 + (b – c)2 – (a – b) (b – c)] Take a-b =p and b-c=q then(p+q)(p2-pq+q2) which is equal to (p+q)3So 3 =(a-c)3 ={(a-c)+(b-c)}(a – c) [(a – b)2 + (b – c)2 – (a – b) (b – c)] So (a – c) [(a – b)2 + (b – c)2 – (a – b) (b – c)] + (c – a)3 = =03+(c-a)3 (a-c)
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