If $\frac{1}{x^3\left(x+3\right)}=\frac{1}{Ax}-\frac{1}{B{x}^2}+\frac{1}{C{x}^3}-\frac{1}{D\left(x+3\right)}$  then $4A+4B+C+Disabc$  $thenb-\left(a+c\right)=$

$1=A{x}^2\left(x+3\right)+Bx\left(x+3\right)+C\left(x+3\right)+D{x}^3$

Comparing both sides

$A+D=0;3A+B=0;3B+C=0;3C=1\Rightarrow C=\frac{1}{3}$

$3B+C=0\Rightarrow B=-\frac{1}{9}$

$3A+B=0\Rightarrow A=\frac{1}{27}$

$A+D=0\Rightarrow D=\frac{-1}{27}$

$\therefore \frac{1}{x^3\left(x+3\right)}=\frac{1}{27x}-\frac{1}{9x^2}+\frac{1}{3x^3}-\frac{1}{27\left(x+3\right)}$

Given that

$\frac{1}{x^3\left(x+3\right)}=\frac{1}{Ax}-\frac{1}{B{x}^2}+\frac{1}{C{x}^3}-\frac{1}{D\left(x+3\right)}$

$A+B+C+D=27+9+3+27=66$

$4A+4B+C+D=108+36+3+27=174$

$thenb-\left(a+c\right)=$ 2