The term independent of x in the binomial expansion of `(1-1/x+3x^(5))(2x^(2)-1/x)^(8)` is:

Options

A. 400
B. 496
C. `-400`
D. `-496`

Answer

To find the term independent of \( x \) in the binomial expansion of \( (1 - \frac{1}{x} + 3x^5)(2x^2 - \frac{1}{x})^8 \), we will follow these steps:

### Step 1: Expand the Binomial Expression
We start with the binomial expansion of \( (2x^2 - \frac{1}{x})^8 \). The general term in the expansion can be expressed using the binomial theorem:

\[ T_{r+1} = \binom{8}{r} (2x^2)^{8-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the General Term
Now, we simplify the general term:

\[ T_{r+1} = \binom{8}{r} (2^{8-r} x^{2(8-r)}) \left(-1\right)^r x^{-r} \]
\[ = \binom{8}{r} (-1)^r 2^{8-r} x^{16 - 3r} \]

### Step 3: Identify the Terms from the First Factor
Next, we consider the first factor \( (1 - \frac{1}{x} + 3x^5) \). We need to find the contributions from each term when multiplied by the general term \( T_{r+1} \).

1. **From \( 1 \)**: The term is \( T_{r+1} \).
2. **From \( -\frac{1}{x} \)**: The term is \( -\frac{1}{x} T_{r+1} \).
3. **From \( 3x^5 \)**: The term is \( 3x^5 T_{r+1} \).

### Step 4: Set Up Conditions for Independence from \( x \)
We need to find the values of \( r \) such that the total exponent of \( x \) in each case is zero.

1. **From \( 1 \)**: \( 16 - 3r = 0 \Rightarrow r = \frac{16}{3} \) (not an integer).
2. **From \( -\frac{1}{x} \)**: \( 16 - 3r - 1 = 0 \Rightarrow 16 - 3r = 1 \Rightarrow r = 5 \).
3. **From \( 3x^5 \)**: \( 16 - 3r + 5 = 0 \Rightarrow 16 - 3r = -5 \Rightarrow r = 7 \).

### Step 5: Calculate the Coefficients
Now we calculate the coefficients for \( r = 5 \) and \( r = 7 \).

1. **For \( r = 5 \)**:
\[ T_{6} = \binom{8}{5} (-1)^5 2^{8-5} = \binom{8}{5} (-1) 2^3 = -56 \cdot 8 = -448 \]

2. **For \( r = 7 \)**:
\[ T_{8} = \binom{8}{7} (-1)^7 2^{8-7} = \binom{8}{7} (-1) 2^1 = -8 \cdot 2 = -16 \]

### Step 6: Combine the Coefficients
Finally, we combine the contributions from both cases:

\[ \text{Total Coefficient} = -448 + 3 \cdot (-16) = -448 - 48 = -496 \]

### Conclusion
The term independent of \( x \) in the binomial expansion is \( -496 \).

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