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• Comment: This content is already covered at Factorization#Recognizable patterns. So, one can use this title for a redirect to this section. Otherwise, creating an article with this title would be a WP:REDUNDANTFORK

• Comment: A redirect has already been created. Unless someone objects in the next few days I will delete

• Comment: This technique needs to demonstrate notability by showing coverage in secondary sources. Further, the derivations need to be sourced so as to not appear to be original research. microbiologyMarcus ^{( petri dish· growths)} 17:36, 12 December 2023 (UTC)

• Comment: I think this idea that this is a duplicate makes sense from the point of view of a mathematician but not from the more enlightened and intelligent point of view of a teacher of mathematics. Michael Hardy ( talk) 04:17, 28 December 2023 (UTC)

• Comment: This content is already covered at Factorization#Recognizable patterns. So, one can use this title for a redirect to this section. Otherwise, creating an article with this title would be a WP:REDUNDANTFORK.
• Comment: A redirect has already been created. Unless someone objects in the next few days I will delete— Preceding unsigned comment added by Ldm1954 ( talk • contribs)

In mathematics, sums and differences of powers are expressions of the form ${\displaystyle x^{n}+y^{n}}$ and ${\displaystyle x^{n}-y^{n}}$, respectively, where ${\textstyle n}$ is a positive integer. They can be factored by a method that is a generalization of the factorization of the difference of squares and the sum of cubes.

Difference of powers

The expression ${\displaystyle x^{n}-y^{n}}$ can be factored by a method concretely exemplified by this case:

$${\displaystyle x^{5}-y^{5}=(x-y)(x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4}).}$$

That pattern works with other powers than the 5th, as follows: ^[1]

$${\displaystyle {\begin{aligned}x^{n}-y^{n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})\\[5pt]&=(x-y)\sum _{k=0}^{n-1}x^{n-k-1}y^{k}\end{aligned}}}$$

where the second one is written in sigma notation.

Table of special cases

The first five cases are as follows:

Difference of powers factorization

n	expression: ${\displaystyle x^{n}-y^{n}}$	factorization
1	${\displaystyle x-y}$	${\displaystyle x-y}$
2	${\displaystyle x^{2}-y^{2}}$	${\displaystyle (x-y)(x+y)}$
3	${\displaystyle x^{3}-y^{3}}$	${\displaystyle (x-y)(x^{2}+xy+y^{2})}$
4	${\displaystyle x^{4}-y^{4}}$	${\displaystyle (x-y)(x^{3}+x^{2}y+xy^{2}+y^{3})}$
5	${\displaystyle x^{5}-y^{5}}$	${\displaystyle (x-y)(x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4})}$

Proof

To show that the factorization is true, expand the expression on the right-hand side: ^{[2] [1]}

$${\displaystyle {\begin{aligned}&(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})\\[5pt]={}&x(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})-y(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})\\[5pt]={}&x^{n}+(x^{n-1}y+x^{n-2}y^{2}+\cdots +x^{2}y^{n-2}+xy^{n-1})-(x^{n-1}y+x^{n-2}y^{2}+\cdots +x^{2}y^{n-2}+xy^{n-1})-y^{n}\\[5pt]={}&x^{n}-y^{n}\end{aligned}}}$$

Use in calculus

Early in a beginning calculus course, one learns the power rule:

$${\displaystyle {\frac {d}{dx}}x^{n}=nx^{n-1}}$$

One way to prove this uses the factorization of powers, illustrated here in the case ${\textstyle n=5:}$

$${\displaystyle {\begin{aligned}{\frac {d}{dx}}x^{5}={}&\lim _{\Delta x\to 0}{\frac {\Delta (x^{5})}{\Delta x}}=\lim _{u\to x}{\frac {u^{5}-x^{5}}{u-x}}\\[8pt]={}&\lim _{u\to x}{\frac {(u-x)(u^{4}+u^{3}x+u^{2}x^{2}+ux^{3}+x^{4})}{u-x}}\\[8pt]={}&\lim _{u\to x}(u^{4}+u^{3}x+u^{2}x^{2}+ux^{3}+x^{4})\\[8pt]={}&x^{4}+x^{3}\cdot x+x^{2}\cdot x^{2}+x\cdot x^{3}+x^{4}\\[8pt]={}&x^{4}+x^{4}+x^{4}+x^{4}+x^{4}\\[8pt]={}&5x^{4}.\end{aligned}}}$$

Sum of odd powers

The expression ${\displaystyle x^{2n+1}+y^{2n+1}}$ can be factored as follows: ^[3]

$${\displaystyle {\begin{aligned}x^{2n+1}+y^{2n+1}&=(x-y)(x^{2n}-x^{2n-1}y+\cdots -xy^{2n-1}+y^{2n})\\[5pt]&=(x-y)\sum _{k=0}^{2n}(-1)^{k}x^{2n-k}y^{k}\end{aligned}}}$$

Table of values

Table of values for ${\displaystyle 1\leq n\leq 5}$:

Sum of odd powers factorization

n	expression: ${\displaystyle x^{2n+1}+y^{2n+1}}$	factorization
1	${\displaystyle x+y}$	${\displaystyle x+y}$
2	${\displaystyle x^{3}+y^{3}}$	${\displaystyle (x+y)(x^{2}-xy+y^{2})}$
3	${\displaystyle x^{5}+y^{5}}$	${\displaystyle (x+y)(x^{4}-x^{3}y+x^{2}y^{2}-xy^{3}+y^{4})}$
4	${\displaystyle x^{7}+y^{7}}$	${\displaystyle (x+y)(x^{6}-x^{5}y+x^{4}y^{2}-x^{3}y^{3}+x^{2}y^{4}-xy^{5}+y^{6})}$
5	${\displaystyle x^{9}+y^{9}}$	${\displaystyle (x+y)(x^{8}-x^{7}y+x^{6}y^{2}-x^{5}y^{3}+x^{4}y^{4}-x^{3}y^{5}+x^{2}y^{6}-xy^{7}+y^{8})}$

Proof

The sum of powers factorization can be derived from the difference of powers factorization. ^{[4] [5]}

$${\displaystyle x^{2n+1}-y^{2n+1}=(x-y)(x^{2n}+x^{2n-1}y+\cdots +xy^{2n-1}+y^{2n})}$$

Let ${\displaystyle y=-z}$ and substitute this expression into the previous equation.

$${\displaystyle x^{2n+1}-(-z)^{2n+1}=(x-(-z))(x^{2n}+x^{2n-1}(-z)+\cdots +x(-z)^{2n-1}+(-z)^{2n})}$$

An negative number raised to an odd exponent will result in a negative number. Similarly, if a negative number is raised to an even exponent, it will result in a positive number. Using such reasoning, one can deduct that ${\displaystyle (x+y)\nmid (x^{2n}+y^{2n}),\forall n\in \mathbb {Z} ^{+}}$.

$${\displaystyle x^{2n+1}+z^{2n+1}=(x+z)(x^{2n}-x^{2n-1}z+\cdots -xz^{2n-1}+z^{2n})}$$

Complex conjugate

Given that ${\displaystyle i^{2}=-1}$, ${\displaystyle a^{2}+b^{2}}$ can be factored using [[complex numbers]s.

$${\displaystyle a^{2}+b^{2}=(a+bi)(a-bi)}$$

Using this property, all expressions in the form ${\displaystyle (x^{2n}+y^{2n})}$ can factored as ${\displaystyle (x^{n}+iy^{n})(x^{n}-iy^{n})}$.

Examples Information

Example 1

Factorization of ${\displaystyle x^{4}+y^{4}}$.

$${\displaystyle {\begin{aligned}&x^{4}+y^{4}=x^{4}+y^{4}-2x^{2}y^{2}+2x^{2}y^{2}\\[8pt]={}&(x^{2}+y^{2})^{2}-2x^{2}y^{2}\\[8pt]={}&(x^{2}+y^{2})^{2}-({\sqrt {2}}xy)^{2}\\[8pt]&{\text{This is then factored as a difference of two squares:}}\\={}&(x^{2}+{\sqrt {2}}xy+y^{2})(x^{2}-{\sqrt {2}}xy+y^{2}).\end{aligned}}}$$

Example 2

Factorization of ${\displaystyle x^{6}+y^{6}}$.

$${\displaystyle {\begin{aligned}{\vphantom {\int }}&x^{6}+y^{6}\\[8pt]={}&(x^{2})^{3}+(y^{2})^{3}\\[8pt]&{\text{Then factor this as a sum of odd powers:}}\\={}&(x^{2}+y^{2})(x^{4}-x^{2}y^{2}+y^{4})\end{aligned}}}$$

References Information

External links

• Sum and difference of powers at Art of Problem Solving
• Difference of powers at themathpage.com