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

${ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \mathrm{y}=135$nC1​xn−1y=135....(i) ${ }^n C_2 x^{n-2} y^2=30$nC2​xn−2y2=30....(ii) ${ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^{\mathrm{n}-3} \mathrm{y}^3=\frac{10}{3}$nC3​xn−3y3=310​....(iii) $\begin{aligned} & \text { By } \frac{(i)}{(\text { ii) }} \\ & \frac{{ }^n C_1}{{ }^n C_2} \frac{x}{y}=\frac{9}{2}...(iv) \\ & \text { By } \frac{(\text { ii) }}{(\text { iii) }} \\ & \frac{{ }^n C_2}{{ }^n C_3} \frac{x}{y}=9....(v)\end{aligned}$​ By ( ii) (i)​nC2​nC1​​yx​=29​...(iv) By ( iii) ( ii) ​nC3​nC2​​yx​=9....(v)​ $\begin{aligned} & \text { By } \frac{(\text { iv })}{(v)} \\ & \frac{{ }^n C_1{ }^n C_3}{{ }^n C_2{ }^n C_2}=\frac{1}{2} \\ & \frac{2 n^2(n-1)(n-2)}{6}=\frac{n(n-1)}{2} \frac{n(n-1)}{2} \\ & 4 n-8=3 n-3 \\ & \Rightarrow n=5 \\ & \text { put in (v) }\end{aligned}$​ By (v)( iv )​nC2​nC2​nC1​nC3​​=21​62n2(n−1)(n−2)​=2n(n−1)​2n(n−1)​4n−8=3n−3⇒n=5 put in (v) ​ $\begin{aligned} & \frac{x}{y}=9 \\ & x=9 y \\ & \text { put in (i) } \\ & { }^5 C_1 x^4\left(\frac{x}{9}\right)=135 \\ & x^5=27 \times 9 \\ & \Rightarrow x=3, \quad y=\frac{1}{3} \\ & 6\left(n^3+x^2+y\right) \\ & =6\left(125+9+\frac{1}{3}\right) \\ & =806\end{aligned}$​yx​=9x=9y put in (i) 5C1​x4(9x​)=135x5=27×9⇒x=3,y=31​6(n3+x2+y)=6(125+9+31​)=806​