 Hello and welcome to the session. I am Asha and I am going to help you with the following question which says if A and B are distinct integers, prove that A minus B is a factor of A raise to the power n minus B raise to the power n whenever n is a positive integer. We have given a hint, write A raise to the power n is equal to A minus B plus B raise to the power n and expand. Let us now begin with the solution and with the help of this end we will solve the problem which will make it very easy. Now as we know A can be written as A plus B minus B or on rearranging it can be written as B plus A minus B therefore A raise to the power n will be equal to B plus of A minus B raise to the power n. On expanding the right hand side with the help of binomial theorem we have on the left hand side A raise to the power n and on the left hand side nC0 B raise to the power n into A minus B raise to the power 0 plus nC1 B raise to the power n minus 1 into A minus B raise to the power 1 plus nC2 B raise to the power n minus 2 into A minus B raise to the power 2 plus nCn B raise to the power 0 into A minus B raise to the power n or it can further be written as A raise to the power n is equal to B raise to the power n plus nC1 B raise to the power n minus 1 into A minus B plus nC2 B raise to the power n minus 2 into A minus B whole square plus goes on up to plus nCn is 1 and B raise to the power 0 is again 1 then we have A minus B raise to the power n. Now taking B raise to the power n on the left hand side we have A raise to the power n minus B raise to the power n is equal to and now taking A minus B common from all the terms we get A minus B and inside the bracket we have nC1 B raise to the power n minus 1 plus nC2 B raise to the power n minus 2 plus goes on up to nCn is 1 and here we have A minus B raise to the power n minus 1 and this shows A minus B is a factor of A minus B raise to the power n this completes the session hope you enjoyed it take care and have a good day.