How do you simplify #((x^2-y^2)(x^2+xy+y^2))/((x^3-y^3)(x^2+2xy+y^2))#?

2 Answers
Apr 15, 2018

Answer:

It simplifies to #1/(x+y)#.

Explanation:

First, factor the bottom right and top left polynomials using the special binomial factoring cases:

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

#=(color(green)((x-y)(x+y))(x^2+xy+y^2))/((x^3-y^3)color(blue)((x+y)(x+y)))#

Cancel the common factor:

#=(color(green)((x-y)color(red)cancelcolor(green)((x+y)))(x^2+xy+y^2))/((x^3-y^3)color(blue)((x+y)color(red)cancelcolor(blue)((x+y))))#

#=(color(green)((x-y))(x^2+xy+y^2))/((x^3-y^3)color(blue)((x+y)))#

Next, use the difference of cubes product to factor the bottom left polynomial:

#=(color(green)((x-y))(x^2+xy+y^2))/(color(magenta)((x^3-y^3))color(blue)((x+y)))#

#=(color(green)((x-y))(x^2+xy+y^2))/(color(magenta)((x-y)(x^2+xy+y^2))color(blue)((x+y)))#

Cancel the common factors again:

#=(color(red)cancelcolor(green)((x-y))color(red)cancelcolor(black)((x^2+xy+y^2)))/(color(magenta)(color(red)cancelcolor(magenta)((x-y))color(red)cancelcolor(magenta)((x^2+xy+y^2)))color(blue)((x+y)))#

#=1/color(blue)(x+y)#

That's as simplified as it gets. Hope this helped!

Apr 15, 2018

Answer:

#1/(x+y)#

Explanation:

I'll use the following formulas:

  • #color(blue)(x^2 - y^2 = (x+y)(x-y))#
  • #color(purple)(x^3 - y^3 = (x-y)(x^2 + xy + y^2))#
  • #color(green)((x+y)^2 = x^2 + 2xy + y^2)#

#( color(blue)((x^2 - y^2 )) (x^2 + xy + y^2)) / (color(purple)((x^3 - y^3)) color(green)((x^2 + 2xy + y^2))#

#=( color(blue)((x+y)(x-y)) (x^2 + xy + y^2)) / (color(purple)((x-y) (x^2 + xy + y^2)) color(green)( (x+y)^2) )#

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

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

#=1/(x+y)#