In the binomial expansion of `(x + y)^n` second, third and forth terms are respectively `135, 30` and `(10)/3` then value of `6(n^2 + x^3 + y^3)`is

Options

A. 2810
B. `(2810)/9`
C. `(1405)/9`
D. `(2810)/9`

Answer

To solve the problem, we need to find the value of \( 6(n^2 + x^3 + y^3) \) given the second, third, and fourth terms of the binomial expansion \( (x + y)^n \) are \( 135, 30, \) and \( \frac{10}{3} \) respectively.

### Step-by-step Solution:

1. **Identify the terms in the binomial expansion**:
The \( r \)-th term in the binomial expansion of \( (x + y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
Therefore, we can express the second, third, and fourth terms as follows:
- Second term \( T_2 = \binom{n}{1} x^{n-1} y = 135 \)
- Third term \( T_3 = \binom{n}{2} x^{n-2} y^2 = 30 \)
- Fourth term \( T_4 = \binom{n}{3} x^{n-3} y^3 = \frac{10}{3} \)

2. **Set up equations based on the terms**:
From the above expressions, we can set up the following equations:
\[ \binom{n}{1} x^{n-1} y = 135 \quad \text{(1)} \]
\[ \binom{n}{2} x^{n-2} y^2 = 30 \quad \text{(2)} \]
\[ \binom{n}{3} x^{n-3} y^3 = \frac{10}{3} \quad \text{(3)} \]

3. **Express \( \binom{n}{r} \)**:
Recall that:
\[ \binom{n}{1} = n, \quad \binom{n}{2} = \frac{n(n-1)}{2}, \quad \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]
Substitute these into the equations:
- From (1): \( n x^{n-1} y = 135 \)
- From (2): \( \frac{n(n-1)}{2} x^{n-2} y^2 = 30 \)
- From (3): \( \frac{n(n-1)(n-2)}{6} x^{n-3} y^3 = \frac{10}{3} \)

4. **Divide equations to eliminate variables**:
Divide equation (2) by equation (1):
\[ \frac{\frac{n(n-1)}{2} x^{n-2} y^2}{n x^{n-1} y} = \frac{30}{135} \]
Simplifying gives:
\[ \frac{(n-1) y}{2 x} = \frac{2}{9} \quad \Rightarrow \quad (n-1) y = \frac{4}{9} x \quad \text{(4)} \]

Now divide equation (3) by equation (2):
\[ \frac{\frac{n(n-1)(n-2)}{6} x^{n-3} y^3}{\frac{n(n-1)}{2} x^{n-2} y^2} = \frac{\frac{10}{3}}{30} \]
Simplifying gives:
\[ \frac{(n-2) y}{3 x} = \frac{1}{9} \quad \Rightarrow \quad (n-2) y = \frac{1}{3} x \quad \text{(5)} \]

5. **Substitute and solve for \( n \)**:
From equation (4), we have:
\[ y = \frac{4}{9(n-1)} x \]
Substitute \( y \) into equation (5):
\[ (n-2) \left(\frac{4}{9(n-1)} x\right) = \frac{1}{3} x \]
Cancel \( x \) (assuming \( x \neq 0 \)):
\[ \frac{4(n-2)}{9(n-1)} = \frac{1}{3} \]
Cross-multiplying gives:
\[ 4(n-2) = 3(n-1) \quad \Rightarrow \quad 4n - 8 = 3n - 3 \quad \Rightarrow \quad n = 5 \]

6. **Find \( x \) and \( y \)**:
Substitute \( n = 5 \) back into equation (4):
\[ (5-1) y = \frac{4}{9} x \quad \Rightarrow \quad 4y = \frac{4}{9} x \quad \Rightarrow \quad y = \frac{1}{9} x \]
Substitute \( y \) into equation (1):
\[ 5 x^{4} \left(\frac{1}{9} x\right) = 135 \quad \Rightarrow \quad \frac{5}{9} x^{5} = 135 \quad \Rightarrow \quad x^{5} = 243 \quad \Rightarrow \quad x = 3 \]
Thus, \( y = \frac{1}{9} \cdot 3 = \frac{1}{3} \).

7. **Calculate \( 6(n^2 + x^3 + y^3) \)**:
Now we can find:
\[ n^2 = 5^2 = 25, \quad x^3 = 3^3 = 27, \quad y^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]
Therefore,
\[ n^2 + x^3 + y^3 = 25 + 27 + \frac{1}{27} = 52 + \frac{1}{27} = \frac{1404 + 1}{27} = \frac{1405}{27} \]
Finally,
\[ 6(n^2 + x^3 + y^3) = 6 \cdot \frac{1405}{27} = \frac{8430}{27} = 310 \]

### Final Answer:
\[ \boxed{310} \]