Question

How to prove (x^n - a^n)/(x-a ) = (x-a)(x^ (n-1) + x^(n-2) a +….a^ (n-1))? Please help.Thanks

Answer 1

\[x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1})\]?

Answer 2

u can use long division and divide \(x^n-a^n\) by \(x-a\) ...

Answer 3

sorry synthetic division

Answer 4

first step\[x^n - a^n-\color\red{ {x^{n-1}}}(x-a)=x^{n-1}a-a^n\]second\[x^{n-1}a-a^n-\color\red{x^{n-2}a} (x-a)=x^{n-2}a^2-a^n\]...until the exponent of a becomes 0

Answer 5

Oops..typo again exponent of x becomes 0

Answer 6

I'll use the same method I'd use to prove it, to make a couple of examples.
(x^3+x^2+x^1+x^0)*(x-1)=
(1*x^4+1*x^3+1*x^2+1*x^1)
-(0*x^4+1*x^3+1*x^2+1*x^1+1*x^0)
=

(1*x^4+1*x^3+1*x^2+1*x^1)
+( -1*x^3-1*x^2-1*x^1-1*x^0)
It's easy to see that x^i=0 to all i other than n and 0, in this example 4 and 0.
Then this is equal to:
x^4-x^0=x^4-1
Second example:
(3^0x^3+3^1x^2+3^2x^1+3^3x^0)(x-3)=
x(3^0x^3+3^1x^2+3^2x^1+3^3x^0)
-3(3^0x^3+3^1x^2+3^2x^1+3^3x^0)
=
(3^0x^4+3^1x^3+3^2x^2+3^3x^1)
- (3^1x^3+3^2x^2+3^3x^1+3^4x^0)
=
(3^0x^4+3^1x^3+3^2x^2+3^3x^1)
+ -3^1x^3-3^2x^2-3^3x^1-3^4x^0)
=
3^0x^4-3^4x^0
=
x^4-3^4
I'll leave the general proof to you, the method is almost the same.

Answer 7

It's like multipling in the primary school.

Answer 8

mukkushla and squeen thanks..@squeen : instead of proving it in reverse way (mukltiplication) i was wondering if there is any direct method w/o having to divide

Answer 9

Dividing a for b is to multiplicate for the inverse multiplicative of a number, when you divide a for b, multiplicating is an operation (I'm praying I'm using correct words because I don't speak English very well), but dividing isn't.
So I think it's a direct proof.

Answer 10

@sqeen :thank you, @mukkshala:thanks again

Answer 11

PD: That's why multiplicate if pretty straight foward, and dividing usually takes longer.

Answer 12

welcome

Answer 13

\[

(x-a)\left (x^ {n-1} + x^{n-2} a +….a^ {n-1}\right)\\
x^n +x^{n-1} a +x^{n-2}a^2\cdots + x a^{n-1}\\
-x^{n-1} a -x^{n-2}a^2\cdots -x a^{n-1}-a^n=x^n - a^n
\]