Asked by  | 11th Feb, 2010, 06:21: PM

Expert Answer:

Let's consider a3 + b3, for example,

(a+b)3 = a3+b3 + 3ab(a+b)

a3+b3= (a+b)3- 3ab(a+b)

(a3+b3)(a+b) = [(a+b)3- 3ab(a+b)]/(a+b) = (a+b)2- 3ab

Similarly for in general using the binomial expansion,

(a+b)n = an + nan-1b + n(n-1)an-2b2/2 ...............................n(n-1)a2bn-2/2 + nabn-1 + bn

But the second and second last term,

nan-1b + nabn-1= n(anb/a + bna/b) = n(anb2 + bna2)/ab = nab(an-2+bn-2)

Now since n is odd, n-2 will be odd, and successive we can reduced the power until we get (a+b). This can be done with all the terms and we can divide by (a+b)

Using the same arguments we can prove the same, for an - bn.




Answered by  | 12th Feb, 2010, 05:34: PM

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