Question

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

Answer 1

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!

Answer 2

`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)`