Simplifying Rational Expressions?

Question

(x^3 - xy^2 )..... ( x^2 + xy ) ... y
------------- ...... x ------------------ x----------
x^2 + 2xy + y^2 ) ..... ( x^2y - xy^3 ) .... x^2


(a^2 b - ab^2 ) .... ( a^2 - b^2 ) ... ....... ... (b^2)
---------------- x ----------------- ----- x -----------
(a + b ) ................. ( ab^2 - b^3) ............... (a^2 - ab)


(a^3 - 3a^2).......... (b^2 - 4) ...... .... ...... (ab - 2a)
----------------- x --------------------- divided by ------------------------
(b^2 + 2b) ...... ........ (a^2 b - 3 ab) ... ... .. .... (b^2 - b^3)


(x^3 + y^3) ..... (x^2 - xy - 6y^2)...... . (x^2-xy+y^2)
------------------------- x ------------------------ divided by-----------------
(x^2 + 3xy + 2y^2) ....... ( x^2 - 2xy - 3y^2) ....... (2x^2 + 2xy).....

to make it clearer., it goes like this:
1.)
x^3-xy^2 / x^2 + 2xy + y^2 times x^2 + xy / x^2y^2 - xy^3 times y/ x^2

2.) a^2b - ab^2 / a + b times a^2 - b^2 / ab^2 - b^3 times b^2 / a^2 - ab

3.) a^3 - 3a^2 / b^2 + 2b times b^2 - 4 / a^2b - 3ab divided by ab - 2a / b^2 - b^3

4.) x^3 + y^3 / x^2 + 3xy + 2y^2 times x^2 - xy - 6y^2 / x^2 - 2xy - 3y^2 divided by x^2 - xy + y^2 / 2x^2 + 2xy

Answer 1

I will do the first one:

take the numerators and break them down by factoring, finding common terms, and factoring the difference of squares.

first numerator becomes:
x(x²-y²)(x(x+y))y = x(x-y)(x+y)x (x+y)y = x²y(x+y)²(x-y)
first denominator is:
(x²+ 2xy +y²)xy(x-y²)x² = (x+y)(x+y)x³y(x-y²) = (x+y)²x³y(x-y²)

now you can cancel whatever is common in the fraction:
..................x²y(x+y)²(x-y)
--------------------------------------------------- =
.................(x+y)²x³y(x-y²)

x's cancel to leave one x on the bottom, y's cancel, (x+y)'s cancel, leaving:
..................(x-y)
-----------------------------------
.................x(x-y²)

you can do the same for the others, factor, combine then cancel.

Answer 2

take the numerators and break them down by factoring, finding common terms, and factoring the difference of squares.

first numerator becomes:
x(x²-y²)(x(x+y))y = x(x-y)(x+y)x (x+y)y = x²y(x+y)²(x-y)
first denominator is:
(x²+ 2xy +y²)xy(x-y²)x² = (x+y)(x+y)x³y(x-y²) = (x+y)²x³y(x-y²)

now you can cancel whatever is common in the fraction:
..................x²y(x+y)²(x-...
------------------------------... =
.................(x+y)²x³y(x-y...

x's cancel to leave one x on the bottom, y's cancel, (x+y)'s cancel, leaving:
..................(x-y)
------------------------------...
.................x(x-y²)

you can do the same for the others, factor, combine then cancel.