The partial fraction decomposition of $\frac{3x^3 - 8x^2 + 10}{(x - 1)^4}$ is:

If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$,then the value of $A+B+C+D+E+F=$

$\frac{4x^2+5}{(x-2)^4} = \frac{A}{(x-2)} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4}$,then $\sqrt{\frac{A}{C} + \frac{B}{C} + \frac{D}{C}} = $

If $\frac{x^4}{(x-1)(x-2)(x-3)}=Ax+B \cdot \frac{1}{x-1}+C \cdot \frac{1}{x-2}+D \cdot \frac{1}{x-3}+E$,then $A+B+C+D+E=$

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

Let $\frac{1}{(x^2-3)^2} = \frac{A_1}{x-\sqrt{3}} + \frac{A_2}{(x-\sqrt{3})^2} + \frac{A_3}{x+\sqrt{3}} + \frac{A_4}{(x+\sqrt{3})^2}$. Then,consider the following statements:
$(i)$ All the $A_i$'s are not distinct
(ii) There exists a pair,$A_p$ and $A_q$ such that $A_p^2 = A_q^2$ $(p \neq q)$
(iii) $\sum_{i=1}^4 A_i = \frac{1}{6}$
(iv) $\sum_{i=1}^4 A_i = 1$
Which one of the following is true?