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

(using only real coefficients)

2 Answers
Καδήρ Κ.
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)#

George C.
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)#

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