How do you factor #x^5-y^5#?
(using only real coefficients)
(using only real coefficients)
2 Answers
Explanation:
We know that :
So let's use this :
Explanation:
Given:
#x^5-y^5#
First note that if
#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
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: