1. ## Algebraic Simplification

a - b - (a^2 / a+b)

(y^3 + 1) / (y^2 - 1) divided by (y^3 - y^2 + y) / (y^2 - 2y + 1)

And perhaps solve for 'y':

y^2 = 3y

2y^2 = y

y / y + 2 + 2 / y^2 - 4 = 1

2. Originally Posted by Ogsta101
(y^3 + 1) / (y^2 - 1) divided by (y^3 - y^2 + y) / (y^2 - 2y + 1)
$\frac{\frac{y^3 + 1}{y^2 - 1}}{\frac{y^3 - y^2 + y}{y^2 - 2y + 1}}$

First, let's get rid of those pesky fractions: Multiply the numerator and denominator by $(y^2 - 1)(y^2 - 2y + 1)$

$= \frac{\frac{y^3 + 1}{y^2 - 1}}{\frac{y^3 - y^2 + y}{y^2 - 2y + 1}} \cdot \frac{(y^2 - 1)(y^2 - 2y + 1)}{(y^2 - 1)(y^2 - 2y + 1)}$
(Pop quiz: Note that we only need to multiply by $(y - 1)^2(y + 1)$, why?)

$= \frac{(y^3 + 1)(y^2 - 2y + 1)}{(y^3 - y^2 + y)(y^2 - 1)}$

Now let's do some factoring:
$= \frac{[(y + 1)(y^2 - y + 1)][(y - 1)^2]}{[y(y^2 - y + 1)][(y + 1)(y - 1)]}$

Now let's do some(!) canceling:
$= y - 1;~y \neq -1, 1$

-Dan

3. Originally Posted by Ogsta101
y^2 = 3y
$y^2 = 3y$

$y^2 - 3y = 0$

$y(y - 3) = 0$

Thus
$y = 0$
or
$y - 3 = 0 \implies y = 3$

-Dan