$\text{Binomial}^3$

$\begin{aligned} \sum_{r=0}^{10} \binom {10}r x^r & = (1+x)^{10} \\ \implies \sum_{r=0}^{10} r^3\binom {10}r x^{r-1} & = \frac d{dx} \left(x \cdot \frac d{dx} \left( x \cdot \frac d{dx} (1+x)^{10} \right) \right) \\ & = \frac d{dx} \left(x \cdot \frac d{dx} \left(10x (1+x)^9 \right) \right) \\ & = \frac d{dx} \left(10x(1+x)^9 + 90x^2(1+x)^8 \right) \\ & = 10(1+x)^9 + 90x(1+x)^8 + 180x(1+x)^8 + 720x^2(1+x)^7 & \small \color{#3D99F6} \text{Putting }x=1 \\ \implies \sum_{r=0}^{10} r^3\binom {10}r & = 10(2^9) + 90(2^8) + 180(2^8) + 720(2^7) \\ & = 166400 = 120500 + 45900 \end{aligned}$

Therefore, $x=\boxed{120500}$ .

I assume $n=10$ ?