# difference of two nth powers (proof)

• Jan 8th 2011, 04:32 PM
Bart
difference of two nth powers (proof)
$a^1 - b^1 = (a - b)$
$a^2 - b^2 = (a-b)(a+b)$
$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
$a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3)$

If you continue with increasing the exponent then you notice the following general rule: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1})$
However, this doesn't proof that this rule is valid for every positive integer exponent. How can you proof that this rule indeed is valid for every positive integer exponent?
• Jan 8th 2011, 04:35 PM
Plato
Quote:

Originally Posted by Bart
However, this doesn't proof that this rule is valid for every positive integer exponent. How can you proof that this rule indeed is valid for every positive integer exponent?

Induction?
• Jan 8th 2011, 05:00 PM
Prove It
No need for induction. Just define $\displaystyle n$ to be a positive integer.

Then $\displaystyle (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + a^2b^{n-3} + a\,b^{n-2} + b^{n-1})$

$\displaystyle = a(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + a^2b^{n-3} + a\,b^{n-2} + b^{n-1}) - b(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + a^2b^{n-3} + a\,b^{n-2} + b^{n-1})$

$\displaystyle = (a^n + a^{n - 1}b + a^{n-2}b^2 + \dots + a^3b^{n - 3} + a^2b^{n - 2} + a\,b^{n - 1}) - (a^{n-1}b + a^{n-2}b^2 + a^{n-3}b^3 + \dots + a^2b^{n-2} + a\,b^{n-1} + b^{n})$

$\displaystyle = a^n - b^n$ (since all the middle terms cancel).

Q.E.D.
• Jan 8th 2011, 05:53 PM
Plato
Quote:

Originally Posted by Prove It
No need for induction. Just define $\displaystyle n$ to be a positive integer.
$\displaystyle = a^n - b^n$ (since all the middle terms cancel).

May I ask you this: without induction how do we know that those “middle terms cancel”
I submit that you do not know that.
Actually this a beautifully simple induction question.
• Jan 8th 2011, 06:53 PM
Prove It
Quote:

Originally Posted by Plato
May I ask you this: without induction how do we know that those “middle terms cancel”
I submit that you do not know that.
Actually this a beautifully simple induction question.

I submit that it is known because it is obvious...
• Jan 9th 2011, 07:18 AM
HallsofIvy
Plato, I see your point but I also see Prove It's. You can compare the "middle terms" term by term and see that they cancel in pairs. Of course, you have to assume that the terms covered by the "..." will do the same.
• Jan 9th 2011, 11:07 AM
wonderboy1953
Question
Quote:

Originally Posted by Bart
$a^1 - b^1 = (a - b)$
$a^2 - b^2 = (a-b)(a+b)$
$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
$a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3)$

If you continue with increasing the exponent then you notice the following general rule: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1})$
However, this doesn't proof that this rule is valid for every positive integer exponent. How can you proof that this rule indeed is valid for every positive integer exponent?

Can this be proven using the binomial theorem?
• Jan 9th 2011, 11:32 AM
snowtea
Quote:

Originally Posted by wonderboy1953
Can this be proven using the binomial theorem?

Sure. Let c = a - b

$a^n - b^n = (b + c)^n - b^n = \sum_{k=0}^n \binom{n}{k}b^kc^{n-k} - b^n
= \sum_{k=0}^{n-1} \binom{n}{k}b^kc^{n-k}
$

$
= c\sum_{k=0}^{n-1} \binom{n}{k}b^kc^{n-k-1}$

Now substitue back in for $c$.
$= (a-b)\sum_{k=0}^{n-1} \binom{n}{k}b^k(a - b)^{n-k-1}$
$= (a-b)\sum_{k=0}^{n-1} \binom{n}{k}b^k
\sum_{l=0}^{n-k-1}\binom{n-k-1}{l}a^lb^{n-k-l-1}$

$= (a-b)\sum_{k=0}^{n-1}\sum_{l=0}^{n-k-1}\binom{n}{k}\binom{n-k-1}{l}a^lb^{n-l-1}$
...
You can keep going, but it seems like the other way is easier :)
• Jan 9th 2011, 11:40 AM
TheCoffeeMachine
• Jan 9th 2011, 11:43 AM
wonderboy1953
Quote:

Originally Posted by TheCoffeeMachine

Nice rule