# Difference between revisions of "Sum and difference of powers"

(odd, it does not work for even) |
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<math>1^3+2^3+3^3+4^3=10^2</math> | <math>1^3+2^3+3^3+4^3=10^2</math> | ||

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+ | ==Factorizations of Sums of Powers== | ||

+ | <math>x^2-y^2=(x-y)(x+y)</math> | ||

+ | |||

+ | <math>x^3-y^3=(x-y)(x^2+xy+y^2)</math> | ||

+ | |||

+ | <math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math> | ||

+ | |||

+ | Note that all these sums of powers can be factorized as follows: | ||

+ | |||

+ | If we have a sum of powers of degree "n", then | ||

+ | |||

+ | (x^n)±(y^n)=(x±y)(x^n+(x^(n-1))y+(x^(n-2))y^2...+y^n) | ||

+ | |||

+ | Note that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial (x+y)^n, except for the fact that the coefficient on each of the terms is 1. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem. | ||

+ | |||

+ | - icecreamrolls8 | ||

+ | |||

==See Also== | ==See Also== | ||

* [[Factoring]] | * [[Factoring]] | ||

* [[Difference of squares]], an extremely common specific case of this. | * [[Difference of squares]], an extremely common specific case of this. | ||

+ | * [[Binomial Theorem]] | ||

{{stub}} | {{stub}} | ||

[[Category:Elementary algebra]] | [[Category:Elementary algebra]] |

## Revision as of 14:12, 5 May 2021

The **sum and difference of powers** are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.

## Contents

## Sums of Odd Powers

## Differences of Powers

If is a positive integer and and are real numbers,

For example:

Note that the number of terms in the *long* factor is equal to the exponent in the expression being factored.

An amazing thing happens when and differ by , say, . Then and

.

For example:

If we also know that then:

## Sum of Cubes

## Factorizations of Sums of Powers

Note that all these sums of powers can be factorized as follows:

If we have a sum of powers of degree "n", then

(x^n)±(y^n)=(x±y)(x^n+(x^(n-1))y+(x^(n-2))y^2...+y^n)

Note that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial (x+y)^n, except for the fact that the coefficient on each of the terms is 1. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.

- icecreamrolls8

## See Also

- Factoring
- Difference of squares, an extremely common specific case of this.
- Binomial Theorem

*This article is a stub. Help us out by expanding it.*