How do you simplify #(x^5+y^5)/(x^3+y^3)#?

1 Answer
George C.
Jan 11, 2018

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

#color(white)((x^5+y^5)/(x^3+y^3)) =x^2+y^2-(x^2y^2)/(x^2-xy+y^2)#

Explanation:

Given:

#(x^5+y^5)/(x^3+y^3)#

Note that #x^n+y^n# is always divisible by #x+y# when #n# is odd.

So we find:

#(x^5+y^5)/(x^3+y^3) = (color(red)(cancel(color(black)((x+y))))(x^4-x^3y+x^2y^2-xy^3+y^4))/(color(red)(cancel(color(black)((x+y))))(x^2-xy+y^2))#

#color(white)((x^5+y^5)/(x^3+y^3)) = (x^4-x^3y+x^2y^2-xy^3+y^4)/(x^2-xy+y^2)#

We can attempt to simplify this some more by separating out a multiple of the denominator from the numerator:

#(x^2-xy+y^2)(x^2+kxy+y^2)#

#=x^4+(k-1)x^3y+(2-k)x^2y^2+(k-1)xy^3+y^4#

So we could choose #k=0# to match the outer terms or #k=1# to match the #x^2y^2# term. Let's choose #k=0# to find:

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

So we can write:

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

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