Rearrange the algebraic expression and use the special product rule \displaystyle {{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}

\displaystyle \frac{{{{x}^{2}}+2xy+{{y}^{2}}-{{z}^{2}}}}{{{{x}^{2}}-{{y}^{2}}-2yz-{{z}^{2}}}}

Factorize out the negative sign and use \displaystyle {{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}} once more.

\displaystyle \frac{{{{{(x+y)}}^{2}}-{{z}^{2}}}}{{{{x}^{2}}-({{y}^{2}}+2yz+{{z}^{2}})}}

\displaystyle \frac{{{{{(x+y)}}^{2}}-{{z}^{2}}}}{{{{x}^{2}}-{{{(y+z)}}^{2}}}}

Factorize once more by using the special product rule \displaystyle {{a}^{2}}-{{b}^{2}}=(a+b)(a-b)

\displaystyle \frac{{\left[ {(x+y)+z} \right]\left[ {(x+y)-z} \right]}}{{\left[ {x-(y+z)} \right]\left[ {x+(y+z)} \right]}}

\displaystyle \frac{{\left( {x+y+z} \right)\left( {x+y-z} \right)}}{{(x-y-z)\left( {x+y+z} \right)}}

Simplify by dividing out the common expression.

\displaystyle \frac{{\left( {x+y-z} \right)}}{{(x-y-z)}}

 

 

This question is from the Textbook “New Syllabus Matheematics 2 7th Edition” publish by Shinglee. It appeared in the  Advance section of chapter 6  Algebraic Fractions and Formulas.  Most of my students looked lost when they saw this questions After giving them a few tips they manage to simplify the expression.

It is important that you memorize the three special algebraic product rules below, although you may be able to derive 1) and 2) without memorizing, I feel that you will be able to do much faster if you memorize.

1. \displaystyle {{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}

2. \displaystyle {{a}^{2}}-2ab+{{b}^{2}}={{(a-b)}^{2}}

3. \displaystyle {{a}^{2}}-{{b}^{2}}=(a+b)(a-b)

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Mathematics Tuition in Woodllands, Johor Bahru, Choa Chu Kang, Yew Tee.