Question 212101
{{{y*z*(z-y)+x*y*(y-x)+x*z*(x-z)}}}
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:
{{{-y*z*(y-z)+x*y*(y-x)+x*z*(x-z)}}}
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:
{{{-y*z*(y-z)+x*y^2 - x^2*y + x^2*z - xz^2}}}
Rearranging the terms for factoring by grouping we get:
{{{-y*z*(y-z)+ red(x*y^2 - xz^2) + green(-x^2*y + x^2*z)}}}
Factoring x from the "red" terms and {{{-x^2}}} from the "green" terms we get:
{{{-y*z*(y-z)+ x*red(y^2 - z^2) + (-x^2)*(y - z)}}}
Factoring the "red" expression as a difference of squares we get:
{{{-y*z*(y-z) + x*(y+z)(y-z) + (-x^2)*(y - z)}}}
Now we have (y-z) as a common factor of each term. We can factor it out:
{{{(y-z)(-y*z + red(x*(y+z)) - x^2)}}}
Multiplying out the "red" expression we get:
{{{(y-z)(-y*z + xy + xz - x^2)}}}
Rearranging the terms for factoring by grouping we get:
{{{(y-z)(red(xz - y*z) + green(-x^2 + xy))}}}
Factoring z from the "red" terms and -x from the "green" terms we get:
{{{(y-z)(z(x - y) + (-x)(x - y))}}}
Now we can factor the (x-y)'s:
{{{(y-z)((x-y)(z-x))}}}
With the Associative Property we can remove the extra parentheses giving:
{{{(y-z)(x-y)(z-x)}}}
And with the Commutative Property we have:
{{{(x-y)(y-z)(z-x)}}}

and we're done!