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

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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

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

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

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