How do you factor x^5-y^5?

(using only real coefficients)

2 Answers
Jul 21, 2017

x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)

Explanation:

We know that :

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

So let's use this :

x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)

Jul 21, 2017

x^5-y^5=(x-y)(x^2+(1/2-sqrt(5)/2)xy+y^2)(x^2+(1/2+sqrt(5)/2)xy+y^2)

Explanation:

Given:

x^5-y^5

First note that if x=y then x^5-y^5 = 0. Hence we can deduce that (x-y) is a factor:

x^5-y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)

We can factor the remaining quartic by making use of its symmetry, expressing it in terms of a quadratic in (x/y+y/x) as follows:

Note that:

(x/y+y/x)^2 = x^2/y^2+2+y^2/x^2

So we find:

x^4+x^3y+x^2y^2+xy^3+y^4

= x^2y^2(x^2/y^2+x/y+1+y/x+y^2/x^2)

= x^2y^2((x/y+y/x)^2+(x/y+y/x)-1)

= x^2y^2((x/y+y/x)^2+(x/y+y/x)+1/4-5/4)

= x^2y^2(((x/y+y/x)+1/2)^2-(sqrt(5)/2)^2)

= x^2y^2((x/y+y/x)+1/2)-sqrt(5)/2)((x/y+y/x)+1/2)+sqrt(5)/2)

= x^2y^2(x/y+(1/2-sqrt(5)/2)+y/x)(x/y+(1/2+sqrt(5)/2)+y/x)

= (x^2+(1/2-sqrt(5)/2)xy+y^2)(x^2+(1/2+sqrt(5)/2)xy+y^2)

So putting it all together, we have:

x^5-y^5=(x-y)(x^2+(1/2-sqrt(5)/2)xy+y^2)(x^2+(1/2+sqrt(5)/2)xy+y^2)