Simplify #(x^2-y^2+z^2+2xz)/(x^2+y^2-z^2-2xy)# by factorizing?

Question

1766 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T04:39:29

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

#(x^2-y^2+z^2+2xz)/(x^2+y^2-z^2-2xy)#

= #(x^2+2xz+z^2-y^2)/(x^2-2xy+y^2-z^2)# regrouping for getting squares

= #((x+z)^2-y^2)/((x-y)^2-z^2)# - now using identity for difference of squares

= #((x+z+y)(x+z-y))/((x-y+z)(x-y-z))# - factorizing

= #((x+y+z)(x-y+z))/((x-y+z)(x-y-z))# - regrouping

= #((x+y+z)cancel((x-y+z)))/(cancel((x-y+z))(x-y-z))#

= #(x+y+z)/(x-y-z)#