Cubic summation

$\begin{aligned} \sum_{n=0}^{123456789} \frac{3n^2+9n+7}{(n^2+3n+2)^3} & = \sum_{n=0}^{123456789} \frac{3(n^2+3n+2)+1}{(n^2+3n+2)^3} \\ & = \sum_{n=0}^{123456789} \left( \frac{3}{(n^2+3n+2)^2} + \frac{1}{(n^2+3n+2)^3} \right) \\ & = \sum_{n=0}^{123456789} \left( \frac{3}{(n+1)^2(n+2)^2} + \left( \frac{1}{(n+1)(n+2)} \right)^3 \right) \\ & = \sum_{n=0}^{123456789} \left( \frac{3}{(n+1)^2(n+2)^2} + \left( \frac{1}{n+1} - \frac{1}{n+2} \right)^3 \right) \\ & = \sum_{n=0}^{123456789} \left( \frac{3}{(n+1)^2(n+2)^2} + \frac{1}{(n+1)^3} - \frac{3}{(n+1)^2(n+2)} + \frac{3}{(n+1)(n+2)^2} - \frac{1}{(n+2)^3} \right) \\ & = \sum_{n=0}^{123456789} \left( \frac{3}{(n+1)(n+2)} \left(\frac{1}{(n+1)(n+2)} - \frac{1}{n+1} + \frac{1}{n+2} \right) + \frac{1}{(n+1)^3} - \frac{1}{(n+2)^3} \right) \\ & = \sum_{n=0}^{123456789} \left( \frac{3}{(n+1)(n+2)} \left(\frac{1}{n+1} - \frac{1}{n+2} - \frac{1}{n+1} + \frac{1}{n+2} \right) + \frac{1}{(n+1)^3} - \frac{1}{(n+2)^3} \right) \\ & = \sum_{n=0}^{123456789} \left(\frac{1}{(n+1)^3} - \frac{1}{(n+2)^3} \right) \\ & = 1 - \frac{1}{123456791^3} = \frac{123456791^3-1}{123456791^3} \end{aligned}$

$\Rightarrow B - A = \boxed{1}$

Notice that $n^2+3n+2 = (n+1)(n+2)$ , and $(n+2)^3-(n+1)^3 = 3n^2+9n+7$ , then you can split the faction into $\frac{(n+2)^3-(n+1)^3}{(n+1)^3(n+2)^3} =\frac{1}{(n+1)^3}-\frac{1}{(n+2)^3}$ Here, when we put sigma in, it would be in the form of $(1-\frac{1}{8})+(\frac{1}{8}-\frac{1}{27})+\dots+(\frac{1}{123456790^3}-\frac{1}{123456791^3}) = 1-\frac{1}{123456791^3}$

use $3n^2+9n+7=(n+2)^3-(n+1)^3$