If the middle term of the expansion of `(x^3/3+3/x)^8` is `5670` then sum of all real values of `x` is equal to (A) `6` (B) `3` (C) `0` (D) `2`

Options

A. 6
B. 8
C. 0
D. 4

Answer

To solve the problem, we need to find the middle term of the expansion of \((\frac{x^3}{3} + \frac{3}{x})^8\) and set it equal to 5670. We will then solve for \(x\) and find the sum of all real values of \(x\).

### Step-by-step Solution:

1. **Identify the middle term in the expansion:**
The general term in the binomial expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
Here, \(n = 8\), \(a = \frac{x^3}{3}\), and \(b = \frac{3}{x}\). The middle term when \(n\) is even is given by \(T_{(n/2) + 1}\), which is \(T_5\).

2. **Calculate the 5th term:**
\[ T_5 = \binom{8}{4} \left(\frac{x^3}{3}\right)^{4} \left(\frac{3}{x}\right)^{4} \]
\[ = \binom{8}{4} \left(\frac{x^{12}}{3^4}\right) \left(\frac{3^4}{x^4}\right) \]
\[ = \binom{8}{4} \frac{x^{12}}{3^4} \cdot \frac{3^4}{x^4} = \binom{8}{4} x^{12 - 4} = \binom{8}{4} x^8 \]

3. **Calculate \(\binom{8}{4}\):**
\[ \binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]
Thus, \(T_5 = 70 x^8\).

4. **Set the middle term equal to 5670:**
\[ 70 x^8 = 5670 \]
\[ x^8 = \frac{5670}{70} = 81 \]

5. **Solve for \(x\):**
\[ x^8 = 81 \implies x^2 = \sqrt{81} = 9 \implies x = \pm 3 \]

6. **Find the sum of all real values of \(x\):**
The real values of \(x\) are \(3\) and \(-3\).
\[ \text{Sum} = 3 + (-3) = 0 \]

### Final Answer:
The sum of all real values of \(x\) is \(0\).