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

$$(x+y)^n$$: $$T_2 = nx^{n-1}y = 135$$, $$T_3 = \binom{n}{2}x^{n-2}y^2 = 30$$, $$T_4 = \binom{n}{3}x^{n-3}y^3 = 10/3$$.

$$T_3/T_2 = \frac{(n-1)y}{2x} = \frac{30}{135} = \frac{2}{9}$$.

$$T_4/T_3 = \frac{(n-2)y}{3x} = \frac{10/3}{30} = \frac{1}{9}$$.

Dividing: $$\frac{2(n-2)}{3(n-1)} = \frac{1/9}{2/9} = 1/2$$. So $$4(n-2) = 3(n-1) \Rightarrow n = 5$$.

$$y/x = \frac{2\times2}{9\times4} = 1/9$$... From $$\frac{4y}{2x} = 2/9$$: $$y/x = 1/9$$.

From $$T_2 = 5x^4y = 135$$: $$5x^4(x/9) = 135 \Rightarrow 5x^5/9 = 135 \Rightarrow x^5 = 243 = 3^5 \Rightarrow x = 3$$. $$y = 1/3$$.

$$6(n^3+x^2+y) = 6(125+9+1/3) = 6(134+1/3) = 6(403/3) = 806$$.

The answer is 806.