tuzkutimonq4

2022-09-10

Advice on simple exercise; dividing rational expression

I'm looking into rational expressions to clarify some doubts that i have. I'm solving the exercises at the end of the chapter and I'm stuck with one of them. This is it:

$$\frac{(x-y{)}^{2}-{z}^{2}}{(x+y{)}^{2}-{z}^{2}}\xf7\frac{x-y+z}{x+y-z}$$

I have been solving several of this type by:

1.- Factoring out the largest common factor if possible.

2.- Factoring.

3.- Removing factors of 1.

In this case, i'm trying to find a way of solving this without expanding the hole thing, but i have not found the way jet.

I start by expanding both square binomials of the dividend.

$$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xy+{y}^{2}-{z}^{2}}\xf7\frac{x-y+z}{x+y-z}$$

Then, multiplying the dividend by the reciprocal of the divisor and rewriting.

$$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xy+{y}^{2}-{z}^{2}}\times \frac{x+y-z}{x-y+z}$$

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

Expanding the expression and grouping i got

$$\frac{{x}^{3}-{x}^{2}(y+z)-x(y-z{)}^{2}+{y}^{3}-{y}^{2}z-y{z}^{2}+{z}^{3}}{{x}^{3}+{x}^{2}(y+z)-x(y-z{)}^{2}-{y}^{3}+{y}^{2}z+y{z}^{2}-{z}^{3}}$$

The problem is that i don't know what to do now. I know that the result is:

$$\frac{x-y-z}{x+y+z}$$

I want to belive that there's an easy way from the beggining (to avoid expanding all the thing) but i would like your help.

I'm looking into rational expressions to clarify some doubts that i have. I'm solving the exercises at the end of the chapter and I'm stuck with one of them. This is it:

$$\frac{(x-y{)}^{2}-{z}^{2}}{(x+y{)}^{2}-{z}^{2}}\xf7\frac{x-y+z}{x+y-z}$$

I have been solving several of this type by:

1.- Factoring out the largest common factor if possible.

2.- Factoring.

3.- Removing factors of 1.

In this case, i'm trying to find a way of solving this without expanding the hole thing, but i have not found the way jet.

I start by expanding both square binomials of the dividend.

$$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xy+{y}^{2}-{z}^{2}}\xf7\frac{x-y+z}{x+y-z}$$

Then, multiplying the dividend by the reciprocal of the divisor and rewriting.

$$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xy+{y}^{2}-{z}^{2}}\times \frac{x+y-z}{x-y+z}$$

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

Expanding the expression and grouping i got

$$\frac{{x}^{3}-{x}^{2}(y+z)-x(y-z{)}^{2}+{y}^{3}-{y}^{2}z-y{z}^{2}+{z}^{3}}{{x}^{3}+{x}^{2}(y+z)-x(y-z{)}^{2}-{y}^{3}+{y}^{2}z+y{z}^{2}-{z}^{3}}$$

The problem is that i don't know what to do now. I know that the result is:

$$\frac{x-y-z}{x+y+z}$$

I want to belive that there's an easy way from the beggining (to avoid expanding all the thing) but i would like your help.

Willie Gilmore

Beginner2022-09-11Added 8 answers

Yuck. You seem to have missed a trick early on. Notice that ${x}^{2}-{y}^{2}=(x+y)(x-y)$. That should help out a lot - before you do anything like expand brackets.

iescabroussexg

Beginner2022-09-12Added 4 answers

Expanding the first term gives

$$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xz-{y}^{2}+{z}^{2}}.$$

Now it easy to see that this is equal to

$$\frac{x-y-z}{x+y+z}.$$

$$\frac{{x}^{2}-2xy+{y}^{2}-{z}^{2}}{{x}^{2}+2xz-{y}^{2}+{z}^{2}}.$$

Now it easy to see that this is equal to

$$\frac{x-y-z}{x+y+z}.$$

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