SOLUTION: Dear Tutor, I would like to know how can I arrive/make: y*z*(z-y)+x*y*(y-x)+x*z*(x-z) into (x-y)*(y-z)*(z-x) I have tried multiplying all factors also tried

SOLUTION: Dear Tutor, I would like to know how can I arrive/make: yz(z-y)+xy(y-x)+xz(x-z) into (x-y)(y-z)(z-x) I have tried multiplying all factors also tried

Question 212101: Dear Tutor,

I would like to know how can I arrive/make:
y*z*(z-y)+x*y*(y-x)+x*z*(x-z)

into

(x-y)*(y-z)*(z-x)

I have tried multiplying all factors also tried to divide by factor:y and than factor:z but that's where I got stuck so I'm not sure if I'm even close to the right method.

Thank you for your time and effort,

Gabi
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

This is a tough one. For one, you have to be good at factoring by grouping. Another key is recognizing that:
z-y = -1(y-z)
y-x = -1(x-y)
x-z = -1(z-x)
Substituting the first of these into the original we get:

Now we have one of the desired factors, (y-z), present. The rest is manipulations with the goal of having (y-z) as a common factor. First we'll multiply out the reset of the expression:

Rearranging the terms for factoring by grouping we get:

Factoring x from the "red" terms and from the "green" terms we get:

Factoring the "red" expression as a difference of squares we get:

Now we have (y-z) as a common factor of each term. We can factor it out:

Multiplying out the "red" expression we get:

Rearranging the terms for factoring by grouping we get:

Factoring z from the "red" terms and -x from the "green" terms we get:

Now we can factor the (x-y)'s:

With the Associative Property we can remove the extra parentheses giving:

And with the Commutative Property we have:

and we're done!
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