difference of two nth powers (proof)

**Bart** · January 8th 2011, 03:32 PM

$a^1 - b^1 = (a - b)$
$a^2 - b^2 = (a-b)(a+b)$
$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
$a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3)$

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?

**Plato** · January 8th 2011, 03:35 PM

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?

**Prove It** · January 8th 2011, 04:00 PM

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.

**Plato** · January 8th 2011, 04:53 PM

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.

**Prove It** · January 8th 2011, 05:53 PM

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...

**HallsofIvy** · January 9th 2011, 06:18 AM

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.

**wonderboy1953** · January 9th 2011, 10:07 AM

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?

**snowtea** · January 9th 2011, 10:32 AM

Originally Posted by wonderboy1953

Can this be proven using the binomial theorem?

Sure. Let c = a - b

$a^n - b^n = (b + c)^n - b^n = \sum_{k=0}^n \binom{n}{k}b^kc^{n-k} - b^n<br /> = \sum_{k=0}^{n-1} \binom{n}{k}b^kc^{n-k}<br />$
$<br /> = c\sum_{k=0}^{n-1} \binom{n}{k}b^kc^{n-k-1}$
Now substitue back in for $c$ .
$= (a-b)\sum_{k=0}^{n-1} \binom{n}{k}b^k(a - b)^{n-k-1}$
$= (a-b)\sum_{k=0}^{n-1} \binom{n}{k}b^k<br /> \sum_{l=0}^{n-k-1}\binom{n-k-1}{l}a^lb^{n-k-l-1}$
$= (a-b)\sum_{k=0}^{n-1}\sum_{l=0}^{n-k-1}\binom{n}{k}\binom{n-k-1}{l}a^lb^{n-l-1}$
...
You can keep going, but it seems like the other way is easier

**TheCoffeeMachine** · January 9th 2011, 10:40 AM

Ruffini's rule.

**wonderboy1953** · January 9th 2011, 10:43 AM

Originally Posted by TheCoffeeMachine

Ruffini's rule.

Nice rule

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