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The problem given in the image is about the expansion of \((x + y)^n\). Specifically, the second, third, and fourth terms of the binomial expansion are provided, and we are asked to find the value of the expression \(6(n^3 + x^2 + y)\).

### Breakdown of the Problem:

1. **Binomial Expansion:**
   The binomial expansion of \((x + y)^n\) is:
   \[
   (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k
   \]
   Where \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) objects from \(n\).

2. **Terms in the Expansion:**
   - The **second term** (corresponding to \(k=1\)) is:
     \[
     \binom{n}{1} x^{n-1} y = n x^{n-1} y
     \]
   - The **third term** (corresponding to \(k=2\)) is:
     \[
     \binom{n}{2} x^{n-2} y^2 = \frac{n(n-1)}{2} x^{n-2} y^2
     \]
   - The **fourth term** (corresponding to \(k=3\)) is:
     \[
     \binom{n}{3} x^{n-3} y^3 = \frac{n(n-1)(n-2)}{6} x^{n-3} y^3
     \]

### Given Information:
- The second term is \(135\).
- The third term is \(30\).
- The fourth term is \(\frac{10}{3}\).

### Setting up the equations:

1. From the second term:
   \[
   n x^{n-1} y = 135
   \]
2. From the third term:
   \[
   \frac{n(n-1)}{2} x^{n-2} y^2 = 30
   \]
3. From the fourth term:
   \[
   \frac{n(n-1)(n-2)}{6} x^{n-3} y^3 = \frac{10}{3}
   \]

### Solving these equations step-by-step will lead to determining \(n\), \(x\), and \(y\), which can be used to evaluate the expression \(6(n^3 + x^2 + y)\).

Let me calculate the results for you.It seems that the system of equations arising from the terms in the binomial expansion is quite complex for a direct symbolic solution. However, we can approach it numerically or simplify the terms further to derive the values of \(n\), \(x\), and \(y\).

Let's proceed with a numerical approach to estimate the values and calculate the desired result.The numerical solution to the system of equations is approximately:
- \( n = 5 \)
- \( x = 3 \)
- \( y = \frac{1}{3} \)

Now, we can substitute these values into the expression \(6(n^3 + x^2 + y)\) to find the final result. Let me calculate that next.The value of the expression \(6(n^3 + x^2 + y)\) is equal to **806**.

Let me know if you would like further details on this solution or if you have any questions!

Here are 5 related questions:
1. How does the binomial theorem apply to general expansions like \((x + y)^n\)?
2. What are binomial coefficients, and how are they calculated?
3. What is the significance of the second, third, and fourth terms in the binomial expansion?
4. Can the binomial expansion be applied to negative or fractional powers?
5. How do the terms in binomial expansion relate to combinatorics?

**Tip:** In problems involving binomial expansions, identifying patterns in the binomial coefficients and powers of terms can simplify complex expressions significantly.

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

Solve a binomial expansion problem where the second, third, and fourth terms are given as 135, 30, and 10/3. Learn how to apply the binomial theorem to solve for n, x, and y, and calculate the expression 6(n^3 + x^2 + y). Suitable for Grades 11-12.

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