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Question

If a and b are distinct integers, prove that a – b is a factor of a n – b n , whenever n is a positive integer. [ Hint: write a n = ( a – b + b ) n and expand]

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Solution

The given expression is ( a n b n ) , and we have to prove that ( ab ) is a factor of it where n is positive integer.

a n can be written as, a n = ( ab+b ) n

a n = ( ab+b ) n = [ ( ab )+b ] n = C n 0 ( ab ) n + C n 1 ( ab ) n1 b+.......+ C n n1 ( ab ) b n1 + C n n b n = ( ab ) n + C n 1 ( ab ) n1 b+.......+ C n n1 ( ab ) b n1 + b n

Taking bnto L.H.S, and taking (a – b) common from R.H.S, we get

( a n b n )=( ab )[ ( ab ) n1 + C n 1 ( ab ) n2 b+........+ C n n1 b n1 ] ( a n b n )=k( ab )

Where, k=[ ( ab ) n1 + C n 1 ( ab ) n2 b+........+ C n n1 b n1 ] is a natural number.

Hence, it is proved that the expression ( ab ) is factor of ( a n b n ) , where n is a natural number.


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