Its important pls answer it fully?

Question

prove that if
a^(n) - b^(n) = (a-b)[(a-b)^(n-1)+nk]
then "k" will come natural for n=any prime number.
Please provide the mathematical proof.

Answer 1

let a = b+c
by binomial theoerm

(b+c)^ n has got n+1 terms and p th tern = nCp b^pc^n-k

all terms except b^n and c^n are having n as factor because n does not get cancelled in nCp as n is prime. additionally all of them except b^n has a factor c as it contains a power of c
add all of them and take n as a factor also c

so (b+c)^n = b^n + c^n + nck
put b+ c = a

a^n = b^n +(a-b)^n+n(a-b) k
a^n-b^n = (a-b)^n + n(a-b)k
= (a-b)((a-b)^n-1 + nk)
QED

Answer 2

For binomial property

(a-b)^n = nc0 a^n - nc1 a^n-1 b + nc2 a^n-2 b^2 - nc3 a^n-3 b^3 + ---------- (-1)^n nc(n-1) a b^n-1 - ncn b^n ----------- (1)
(a+b)^n = nc0 a^n + nc1 a^n-1 b + nc2 a^n-2 b^2 + nc3 a^n-3 b^3 + ---------- nc(n-1) a b^n-1 + ncn b^n ---------------(2)

Now subtract (2) - (1)and take your inputs. After simplefication we get the result

a^n - b^n = (a-b) [(a-b)^n-1 + nk]

Answer 3

You said it was important. Did you wanted a dark magician for answering this kind of question? Huh.......

Answer 4

after some trivial manipulations i get :

nk (a-b) = a^(n) - b^(n) - (a-b)^n

since a-b | a^(n) - b^(n) - (a-b)^n

k is natural since n is and lhs is.

now go after the girls.

Answer 5

umhh sorry i am poor in mathematics

Answer 6

You said it was important!!!

Answer 7

yeah u wasted my time too!

Answer 8

do your own homework

Answer 9

D-UH!