Draft:Factorization of sums and differences of powers

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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— Preceding unsigned comment added by Ldm1954 (talk • contribs)

In mathematics, sums and differences of powers are expressions of the form $x^{n}+y^{n}$ and $x^{n}-y^{n}$, 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 $x^{n}-y^{n}$ can be factored by a method concretely exemplified by this case:

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

In the second factor, the powers of x are 4, 3, 2, 1, 0 and the powers of y are 0, 1, 2, 3, 4.

That pattern works with other powers than the 5th, as follows:

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

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

Proof

To show that the factorization is true, expand the expression on the right-hand side:

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

Use in calculus

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

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

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

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

Sum of odd powers

The expression $x^{2n+1}+y^{2n+1}$ can be factored as follows:

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

Table of values

Table of values for $1\leq n\leq 5$:

Sum of odd powers factorization

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

Proof

The sum of powers factorization can be derived from the difference of powers factorization.

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

Let $y=-z$ and substitute this expression into the previous equation.

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

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 $(x+y)\nmid (x^{2n}+y^{2n}),\forall n\in \mathbb {Z} ^{+}$.

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

Complex conjugate

See also: Complex conjugate

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

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

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

Examples

Example 1

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

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

Example 2

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

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

References

External links

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