Question 40707
Simplify.
{{{((y-z)/(a+b))*((a^2-b^2)/(z-y))}}}
First we need to simplify by factoring in order to cancel these expressions.
The numerator in the fraction on the right is the difference of two squares.
{{{a^2 - b^2}}}
We can factor this as (a+b)(a-b)because if we FOIL these out we get {{{a^2 - b^2}}} 
In the denominator of the same fraction we can factor out a -1 in order to make the expression read (y-z) which is NOT the same as (z-y).
So now we have...
{{{((y-z)/(a+b))*((a+b)(a-b)/-1(y-z))}}}
Now we can cancel out all of our like terms.
(y-z) in the numerator of the first expression can cancel with (y-z) in the denominator of the second expression.
{{{((0)/(a+b))*((a+b)(a-b)/-1)}}}
(a+b)in the denominator of the first expression can cancel with (a+b) in the numerator of the second expression.
Which becomes...
{{{(a-b)/-1}}}
Or...
-(a-b)
or...
(-a+b)
or...
(b-a)
This is in its simplist form.
I hope I explained this thoroughly enough for you.
Good Luck!