Please complete the formula a^n-b^n = (a-b) [a^(n-1)+........+b^(n-1)]. How is it derived?

Answer 1

a^n - b^n = ( a - b ) [ a^(n -1) + a^(n - 2)b + a^(n - 3)b^2 + ..... + a^(n - r)b^(r - 1) + ..... + ab^(n - 2) + b^(n - 1) ]

Because if you try to divide a^n - b^n by a - b using long division
(fill out the zeroes between a^n and b^n, i.e. a^n ___ ___ ___ ... ___ b^n),
it will yield the long factor.

Hope this helps you!^_^
^_^

Answer 2

The solution is trivial once you understand the symmetric distribution of the powers of a and b in the binomial expansion. For example:

a^7 - b^7 = (a-b)[a^6 + a^5 b + a^4 b^2 + a^3 b^3 + a^2 b^4 + a b^5
+b^6)

Consider an arbitrary term within the square brackets (excluding a^6). For example, consider the 2nd term which is a^5 b. If we multiply this term by a, then we get a^6 b. But if we multiply the previous term, i.e. a^6, by negative b, then we get -a^6 b and these two terms cancel. Similarly, in general, when we multiply through by (a-b), every term we obtain in mutiplying through by a, is canceled when we multiply the previous term by -b, excluding, of course, the first term, which will be a^n. Thus, the only uncanceled terms will be a^n and -b^n. The cubic example below should make this clear.

a^3 - b^3 = (a-b)[a^2 + ab + b^2], which when expanded

= (a^3 +a^2 b +ab^2) +(-b a^2 - a b^2 -b^3)

= a^3 + (a^2 b - ba^2 + ab^2 - ab^2) - b^3 = a^3 - b^3

The general solution is:

a^n - b^n = (a-b) [a^(n-1) + a^(n-2) b + a^(n-3) b^2+....+a^2 b^(n-3)

+ a b^(n-2) + b^(n-1)]

We see that the solution is easy to construct and has nothing to do with binomial coefficients.

Answer 3

a^n-b^n = (a-b) [a^(n-1)+ a^(n-2)*b +a^(n-3)*b^2 + ....... + a*b^(n-2) +b^(n-1)]

Answer 4

to derive this formula you can use binomial theorem and mathematical induction. the derivation is long but its available in all books related to above topic.

Answer 5

a^n-b^n = (a-b) [a^(n-1)+ (a^(n-2) x b^(n-2)) +b^(n-1)].

It is derived using binomial theorem.......

Answer 6

use binomial theorem and mathematical induction.