The sum of the real values of x for which the middle term in the binomial expansion of `(x^3/3 + 3/x)^8` equals 5670 is :

Options

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

Answer

To solve the problem, we need to find the sum of the real values of \( x \) for which the middle term in the binomial expansion of \( \left( \frac{x^3}{3} + \frac{3}{x} \right)^8 \) equals 5670.

### Step-by-Step Solution:

1. **Identify the Middle Term**:
The binomial expansion of \( (a + b)^n \) has its middle term given by \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \). Here, \( n = 8 \), so the middle term is \( T_{5} \) (since \( r = \frac{n}{2} = 4 \)).

2. **Set Up the Middle Term**:
In our case, \( a = \frac{x^3}{3} \) and \( b = \frac{3}{x} \). Therefore, the middle term \( T_5 \) can be expressed as:
\[ T_5 = \binom{8}{4} \left( \frac{x^3}{3} \right)^{4} \left( \frac{3}{x} \right)^{4} \]

3. **Calculate \( T_5 \)**:
We calculate \( T_5 \):
\[ T_5 = \binom{8}{4} \left( \frac{x^3}{3} \right)^{4} \left( \frac{3}{x} \right)^{4} = \binom{8}{4} \cdot \frac{x^{12}}{3^4} \cdot \frac{3^4}{x^4} \]
Simplifying this gives:
\[ T_5 = \binom{8}{4} \cdot x^{12 - 4} = \binom{8}{4} \cdot x^8 \]

4. **Set the Middle Term Equal to 5670**:
We know from the problem statement that \( T_5 = 5670 \):
\[ \binom{8}{4} \cdot x^8 = 5670 \]

5. **Calculate \( \binom{8}{4} \)**:
The value of \( \binom{8}{4} \) is:
\[ \binom{8}{4} = \frac{8!}{4!4!} = 70 \]
Therefore, we have:
\[ 70 \cdot x^8 = 5670 \]

6. **Solve for \( x^8 \)**:
Dividing both sides by 70:
\[ x^8 = \frac{5670}{70} = 81 \]

7. **Find \( x \)**:
Taking the eighth root of both sides:
\[ x = 3 \quad \text{or} \quad x = -3 \]

8. **Sum of Real Values of \( x \)**:
The real values of \( x \) are \( 3 \) and \( -3 \). Therefore, the sum of these values is:
\[ 3 + (-3) = 0 \]

### Final Answer:
The sum of the real values of \( x \) for which the middle term in the binomial expansion equals 5670 is \( \boxed{0} \).