Question

Prove that : \[\large{x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + y^{n-1}) }\]

Answer 1

The above is for , n is either even or odd.

And for : n is odd can b be written as :
\[\large{x^n + y^n = (x+y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 -...+ y^{n-1})}\]

Answer 2

Yeah, so try multiplying out the LHS.
The last term is still wrong

Answer 3

\[\large{x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + y^{n-1}) }\]

Answer 4

Yeah, now multiply out the brackets. You'd see an interesting pattern and cancellation ;-)

Answer 5

[guys please delete your comments , that are related to the correction of the quest. , this post got long]

Answer 6

Sorry, Parth . Do you have any other method?

Answer 7

use mathematical induction to prove the identity

Answer 8

\[\huge (x+y)^n=\sum_{i=0}^{n}\left(\begin{matrix}n \\ i\end{matrix}\right) x^iy^{n-i}\]

Answer 9

@helder_edwin some more explaination, please.

Answer 10

u can prove the identity you're giveng using mathematical induction.
do u know about it?

Answer 11

*given

Answer 12

Hmn, I will try that.

Answer 13

it should be pretty straightforward

Answer 14

\[\left(x\times x^{n - 1}\right) - \left(y \times x^{n - 1}\right) + \left(x \times x^{n - 2} y\right) \cdots \left( x\times xy^{n - 2}\right) - \left(y\times y^{n-1}\right)\]Take a look at term 2 and 3. They get cancelled. The fourth and the fifth terms also get cancelled. This continues. Though you are left with two terms, i.e., \(x \times x^{n - 1}\) and \(- (y \times y^{n-1})\).

Answer 15

Got it? :-D

Answer 16

Oh, and credits to my teacher. :-)

Answer 17

\(\begin{array}{}
\large{
x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y +... + y^{n-1})\\

\text{Putting n =1} \\
x^1 - y^1 = (x-y)(x^{1-1} + x^{1-2}y + ... + y^{1-1})\\
\implies x - y = (x-y)(x^0 + x^{-1} y + ... +y^{0}) \\
\implies x-y = (x-y)(1 + \cfrac{y}{x} + ... + 1)

}\end{array}\)
I am confused @helder_edwin , I know that x and y must have powers of whole numbers , but pls tell me is it natural numbers ? or whole numbers?

Answer 18

I can delete the following terms : \(\cfrac{y}{x} + ...\) and

\(\implies x-y = (x-y)(1 + 1)\)

But I think the powers of x and y can not be zero .

Answer 19

actually u cannot use whole numbers. the identity is valid only for natural (positive integer) numbers.

Answer 20

Oh k . Thanks a lot helder , so let me continue it

Answer 21

u have to prove
\[ \large x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\dotsb+xy^{n-2}+y^{n-1}) \]

Answer 22

so for \(n=1\) u have
\[ \large x^1-y^1=x-y \]
on the LHS; and
\[ \large (x-y)(x^0+y^0)=(x-y)(1+1)\neq x-y \]

Answer 23

on the RHS.
what do u think of this?
what do u infer about \(n\)?

Answer 24

Well, \(n\ne 0\) , right?

Answer 25

yes. what else?

Answer 26

RHS , I have : \((x-y)(1) = x-y\)

Answer 27

So , we can not write : \((x-y)(x^0 + ... + y^0)\) as \(n \ne 0 \) , thus : \(x-y... (LHS)= (x-y)(1) ...(RHS)\)

Answer 28

LHS = RHS .

Answer 29

it has to be that
\[ \large n\geq2 \]

Answer 30

Oh, so \(n\ne 1\) also. Didn't know that!

Answer 31

Okey, helder so next step will be , trying for n = k and then n = k + 1

Answer 32

[We will assume that the given identity holds true for n = k ] and then try for n = k +1

Answer 33

yes

Answer 34

\[\begin{array}{}
\large{

\text{If n = k is true for the given identity} \\

x^k - y^k = (x-y)(x^{k-1} + x^{k-2} y + ... + y^{k-1})\\
\textbf{Putting n= k +1} \\
x^{k+1} - y^{k+1} = (x-y)(x^{k} + x^{k-1} + ...+ y^{k} )\\


}\end{array}\]

Answer 35

But how will I prove for n = k + 1.

Answer 36

Will I use binomial theorem here ?

Answer 37

it a trick u must have used before. u have
\[ \large x^{k+1}-y^{k+1}=x^{k+1}-y^{k+1}+A-A \]
choose A so that \(x^k-y^k\) can be factored from the RHS

Answer 38

Okey, got it. Thanks a lot Helder [ I have to go away now. ]

Answer 39

glad to help