1. Inequation

I don't know what was done to get the line starting with i.e.

2. $\displaystyle (x-1)(3-x)^2 > x(3-x)(x-1)^2$

$\displaystyle (x-1)(3-x)^2 - x(3-x)(x-1)^2 > 0$

pull out the common factors ...

$\displaystyle (x-1)(3-x)[(3-x) - x(x-1)] > 0$

3. I don't know how the next line was made.

4. For any numbers a and b, $\displaystyle a^2- b^2= (a- b)(a+ b)$

$\displaystyle 3= (\sqrt{3})^2$ so $\displaystyle 3- x^2= (\sqrt{3})^2- x^2= (\sqrt{3}- x)(\sqrt{3}+ x)$

I think that is what you are refering to but, of course, I can't be certain what you mean by "the next line".

5. I understand that. I don't understand how the second line becomes the 3rd line.

6. $\displaystyle (x-1)(3-x)^2 > x(3-x)(x-1)^2$

$\displaystyle (x-1)(3-x)^2 - x(3-x)(x-1)^2 > 0$

pull out the common factors ...

$\displaystyle (x-1)(3-x)[(3-x) \textcolor{red}{- x(x-1)}] > 0$

distribute ...

$\displaystyle (x-1)(3-x)[3-x \textcolor{red}{- x^2 + x}] > 0$

combine like terms ...

$\displaystyle (x-1)(3-x)(3-x^2) > 0$

7. Thanks. I was misreading the equation.