If the last term of `(2^(1//3) -(1)/(sqrt(2)))^(n)` is `((1)/(3.9^(1//3)))^(log_(3)8)`, then the 5th term from the beginning is

Options

A. `""^(10)C_(6)`
B. `(1)/(2)""^(10)C_(4)`
C. `2""^(10)C_(4)`
D. none

Answer

To solve the problem, we need to find the 5th term from the beginning of the expansion of \((2^{1/3} - \frac{1}{\sqrt{2}})^n\) given that the last term is \(\left(\frac{1}{3 \cdot 9^{1/3}}\right)^{\log_3 8}\).

### Step-by-Step Solution:

1. **Identify the Last Term**:
The last term (T_{n+1}) in the binomial expansion of \((a - b)^n\) is given by:
\[ T_{n+1} = \binom{n}{n} a^{n} (-b)^{0} = a^n \]
In our case:
\[ a = 2^{1/3}, \quad b = \frac{1}{\sqrt{2}} \]
So,
\[ T_{n+1} = (2^{1/3})^n = 2^{n/3} \]

2. **Set Up the Equation**:
We know that the last term is equal to:
\[ T_{n+1} = \left(\frac{1}{3 \cdot 9^{1/3}}\right)^{\log_3 8} \]
We can simplify \(9^{1/3}\) as \(3^{2/3}\), hence:
\[ 3 \cdot 9^{1/3} = 3 \cdot 3^{2/3} = 3^{5/3} \]
Therefore,
\[ T_{n+1} = \left(\frac{1}{3^{5/3}}\right)^{\log_3 8} = 3^{-5/3 \cdot \log_3 8} \]

3. **Use Properties of Logarithms**:
We know that \(\log_3 8 = \log_3 (2^3) = 3 \log_3 2\), thus:
\[ T_{n+1} = 3^{-5 \cdot \log_3 2} = 2^{-5} \]

4. **Equate the Two Expressions**:
Now we have:
\[ 2^{n/3} = 2^{-5} \]
This implies:
\[ \frac{n}{3} = -5 \implies n = -15 \]

5. **Find the 5th Term**:
The 5th term \(T_5\) is given by:
\[ T_5 = \binom{n}{4} a^{n-4} (-b)^4 \]
Plugging in our values:
\[ T_5 = \binom{-15}{4} (2^{1/3})^{-15 - 4} \left(-\frac{1}{\sqrt{2}}\right)^4 \]

6. **Calculate Each Component**:
- The binomial coefficient \(\binom{-15}{4}\) can be calculated using the formula for negative integers:
\[ \binom{-n}{r} = (-1)^r \binom{n + r - 1}{r} \]
Thus,
\[ \binom{-15}{4} = (-1)^4 \binom{-15 + 4 - 1}{4} = \binom{-12}{4} \]
- The powers:
\[ (2^{1/3})^{-19} = 2^{-19/3}, \quad \left(-\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} \]

7. **Combine Everything**:
Finally, we combine all the components to find \(T_5\):
\[ T_5 = \binom{-12}{4} \cdot 2^{-19/3} \cdot \frac{1}{4} \]

### Final Answer:
The 5th term from the beginning is:
\[ T_5 = \binom{-12}{4} \cdot 2^{-19/3} \cdot \frac{1}{4} \]