 Hi and welcome to the session. I am Arsha and I am going to help you with the following question which says evaluate i raised to the power 18 plus 1 upon i raised to the power 25 whole raised to the power 3. Now as we know i raised to the power 2 is equal to minus 1 therefore for any integer k i raised to the power 4k is equal to 1 i raised to the power 4k plus 1 is equal to i tau. So, these are some ideas which we are going to use to evaluate a given problem. So, these are our key ideas. Let us now begin with the solution and we have to evaluate i raised to the power 18 plus 1 upon i raised to the power 25 whole raised to the power 3 which can further be written as i raised to the power 18 plus 1 raised to the power 25 on i raised to the power 25 whole raised to the power 3. Now first we will evaluate i raised to the power 18 and then we will evaluate i raised to the power 25. Now i raised to the power 18 can be written as i raised to the power 4 into 4 plus 2 which is equal to i raised to the power 4 into 4 into i raised to the power 2. Now this is in the form of i raised to the power 4k whose value is 1 by a key idea. So, we have 1 into i square and the value of i square is minus 1. Now let us evaluate i raised to the power 25. Now this can be written as i raised to the power 6 into 4 plus 1 that is i raised to the power 6 into 4 into i raised to the power 1 and that is where the key idea is equal to 1 into i just further equal to i. And now let us substitute the values of i raised to the power 18 and i raised to the power 25 we get minus 1 1 raised to the power 25 as 1 upon i raised to the power 25 always to the power 3. Now let us apply the formula of a plus b whole q which is a q plus 3 a square b plus 3 a b square plus b q. Therefore this can further be written as minus 1 whole q plus 3 into minus 1 whole square into 1 upon i eta plus 3 into minus 1 into 1 upon i eta whole square plus 1 upon i eta whole q which is further equal to minus 1 whole cube is minus 1 plus minus 1 whole square is 1 so we have 3 upon i eta and plus into minus is minus minus 3 upon i eta square plus 1 upon i eta cube which can further be written as minus 1 plus 3 upon i eta minus 3 upon i eta square is minus 1 and we have plus i eta cube can be written as i eta into i eta square and i eta square is minus 1 so we have minus i eta. This can further be written as minus 1 plus 3 upon i eta and minus into minus is plus so we have minus 3 minus 1 upon i eta. Now 3 minus 1 is 2 and taking 1 upon i common we have 3 minus 1 so we have 2 plus 2 upon i eta. Now this can further be written as 2 plus 2 into 1 upon i eta and 1 upon i eta can be written as now 1 can be written as minus 1 into minus 1 upon i eta minus 1 is i eta square so writing it i eta square upon i eta now 1 i eta cancels with 1 i eta in the numerator and we have left it minus i eta. Thus this can further be written as 2 plus 2 into minus i eta which is further equal to 2 minus 2 i eta and thus when evaluating we get our answer as 2 minus 2 i eta. So this completes the solution hope you enjoyed it take care and have a good day.