Cross your limits once more

To find the limit $\displaystyle l = \lim_{x \to \infty} \frac{\sum_{n=1}^x n^2(x-n-1)}{\sum_{n=1}^x n^3} = \lim_{x \to \infty} \frac {a_x}{b_x}$ , where $0 < b_1<b_2<...<b_x<...$ and $\displaystyle \lim_{x \to \infty} b_x = \infty$ , we can use Stolz–Cesàro theorem 1 (the $\infty / \infty$ case). That is, if $\displaystyle \lim_{x \to \infty} \frac {a_{x+1}-a_x}{b_{x+1}-b_x} = l \in \mathbb R$ , then $\displaystyle \lim_{x \to \infty} \frac {a_x}{b_x}$ exists and equals to $l$ .

$\begin{aligned} l & = \frac {a_{x+1}-a_x}{b_{x+1}-b_x} \\ & = \lim_{x \to \infty} \frac{\sum_{n=1}^{x+1} n^2(x+1-n-1)-\sum_{n=1}^x n^2(x-n-1)}{\sum_{n=1}^{x+1} n^3 - \sum_{n=1}^x n^3} \\ & = \lim_{x \to \infty} \frac {\sum_{n=1}^{x+1} n^2(x-n)-\sum_{n=1}^x n^2(x-n) + \sum_{n=1}^x n^2}{(x+1)^3} \\ & = \lim_{x \to \infty} \frac {(x+1)^2(x-x-1) + \frac 16 x(x+1)(2x+1)}{(x+1)^3} \\ & = \lim_{x \to \infty} \left(-\frac 1{x+1} + \frac {2x^2+x}{6(x^2+2x+1)} \right) \\ & = \lim_{x \to \infty} \left(-\frac 1{x+1} + \frac {2+\frac 1x}{6\left(1+ \frac 2x+ \frac 1{x^2} \right)} \right) \\ & = \frac 13 \end{aligned}$

$\implies 2a+3b = 2+9 = \boxed{11}$