Simplifying expression help please!!!, these are so hard

e.   To make factoring easier, rearrange the 25 - 20x + 4x^{2 .}

     $\frac{6x^2-11x-10}{6x^2-5x-6}\cdot\frac{6-4x}{4x^2-20x+25}$

Now we need to factor all the numerators and denominators. Start by splitting their middle terms.

=   $\frac{6x^2+4x-15x-10}{6x^2-9x+4x-6}\cdot\frac{6-4x}{4x^2-10x-10x+25}$

                                                                            Now factor by grouping.

=   $\frac{2x(3x+2)-5(3x+2)}{3x(2x-3)+2(2x-3)}\cdot\frac{6-4x}{2x(2x-5)-5(2x-5)}$

=   $\frac{(3x+2)(2x-5)}{(2x-3)(3x+2)}\cdot\frac{-2(2x-3)}{(2x-5)(2x-5)}$

                                                                           Multiply the fractions together.

=   $\frac{(3x+2)(2x-5)(-2)(2x-3)}{(2x-3)(3x+2)(2x-5)(2x-5)}$

                                                                           Cancel the common terms.

=   $\frac{{\color{red}(3x+2)}{\color{red}(2x-5)}(-2){\color{red}(2x-3)}}{{\color{red}(2x-3)(3x+2)(2x-5)}(2x-5)}$

=   $\frac{(-2)}{(2x-5)}$

=   - $\frac{2}{2x-5}$

And  x  ≠  3/2  or  -2/3   since these cause a zero in the denominator of the original expression.

f)

$\dfrac{3x^3-3a^2x}{x^2-2ax+a^2}\cdot\dfrac{a-x}{a^3x+a^2x^2}\\ =\dfrac{3x(x-a)(x+a)}{(x-a)(x-a)}\cdot\left(-\dfrac{x-a}{a^2x(a+x)}\right)\\ =\dfrac{-3x(x-a)^2(x+a)}{a^2x(x-a)^2(x+a)}\\ =-\dfrac{3}{a^2}$