If the second, third and fourth terms in the expansion of $(x+y)^n$ are 135, 30 and $\frac{10}{3}$, respectively, then $6\left(n^3+x^2+y\right)$ is equal to __________.

<p>$$\begin{aligned} & T_2={ }^n C_1 y^1 \cdot x^{n-1}=135 \\ & T_3={ }^n C_2 y^2 \cdot x^{n-2}=30 \\ & T_4={ }^n C_3 y^3 x^{n-3}=\frac{10}{3} \end{aligned}$$</p> <p>$$\begin{aligned} \Rightarrow \frac{135}{30} & =\left(\frac{x}{y}\right) \frac{n \cdot 2}{n(n-1)}=\left(\frac{2}{n-1}\right)\left(\frac{x}{y}\right) \quad \text{... (i)}\\ \frac{30}{\frac{10}{3}} & =\frac{n(n-1)}{2} \frac{3!}{n(n-1)(n-2)}\left(\frac{x}{y}\right) \\ 9 & =\left(\frac{3}{n-2}\right)\left(\frac{x}{y}\right) \end{aligned}$$</p> <p>$$\begin{aligned} \Rightarrow & 3(n-2)=\frac{135}{60}(n-1) \Rightarrow n=5 \\ \Rightarrow & x=9 y \quad \text{.... (i)}\\ & y \cdot x^4=27 \Rightarrow \frac{x}{9} \cdot x^4=3^3 \\ \Rightarrow & x^5=3^5 \Rightarrow x=3 y=\frac{1}{3} \\ \Rightarrow & 6\left(5^3+3^2+\frac{1}{3}\right)=6\left(125+9+\frac{1}{3}\right) \end{aligned}$$</p> <p>$=6(134)+2=806$</p>