If $\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{2x-1}+\frac{B}{3x+1}$,then $16A+9B$ is equal to

Given $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$,find the value of $4A + 2B + 4C$.

If the equivalent partial fraction of $\frac{x^3}{(2 x-1)(x+2)(x-3)}$ is of the form $A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$,then the value of $A+B+C=$

$\begin{aligned} & \frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow A+C= \end{aligned}$

Let $a, b$, and $c$ be such that $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$. Then $\frac{a}{1} + \frac{b}{3} + \frac{c}{5}$ is equal to:

Let $H(x)=3x^4+6x^3-2x^2+1$ and $g(x)$ be a polynomial of degree one. If $\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$, then $H(-1)+2H(2)-3H(1)=$