The term independent of $$x$$ in the binomial expansion of $$\left(1 - \frac{1}{x} + 3x^5\right)\left(2x^2 - \frac{1}{x}\right)^8$$ is

We need to find the term independent of $$ x $$ in the expansion of $$ \left(1 - \frac{1}{x} + 3x^5\right)\left(2x^2 - \frac{1}{x}\right)^8 $$. This term will have an exponent of zero for $$ x $$. We start by expanding the second part using the binomial theorem.

The binomial expansion of $$ \left(2x^2 - \frac{1}{x}\right)^8 $$ is given by:

$$ \left(2x^2 - \frac{1}{x}\right)^8 = \sum_{k=0}^{8} \binom{8}{k} (2x^2)^{8-k} \left(-\frac{1}{x}\right)^k $$

Simplify each term:

$$ \binom{8}{k} (2x^2)^{8-k} \left(-\frac{1}{x}\right)^k = \binom{8}{k} 2^{8-k} x^{2(8-k)} \cdot (-1)^k x^{-k} = \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 2k - k} = \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 3k} $$

So, the general term is $$ T_k = \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 3k} $$.

Now, multiply this by the first part $$ \left(1 - \frac{1}{x} + 3x^5\right) $$:

$$ \left(1 - \frac{1}{x} + 3x^5\right) \sum_{k=0}^{8} T_k = \left(1 - \frac{1}{x} + 3x^5\right) \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} (-1)^k x^{16 - 3k} $$

To find the constant term, we look for combinations where the total exponent of $$ x $$ is zero. This happens when:

1. Multiplying by 1 (exponent 0) and the term from the expansion has exponent $$ 16 - 3k = 0 $$.
2. Multiplying by $$ -\frac{1}{x} $$ (exponent -1) and the term from the expansion has exponent such that $$ (16 - 3k) + (-1) = 0 $$.
3. Multiplying by $$ 3x^5 $$ (exponent 5) and the term from the expansion has exponent such that $$ (16 - 3k) + 5 = 0 $$.

Solve each equation for integer $$ k $$ (since $$ k $$ must be an integer between 0 and 8):

1. $$ 16 - 3k = 0 $$ → $$ 3k = 16 $$ → $$ k = \frac{16}{3} \approx 5.333 $$, not an integer. No solution.
2. $$ 16 - 3k - 1 = 0 $$ → $$ 15 - 3k = 0 $$ → $$ 3k = 15 $$ → $$ k = 5 $$, which is an integer and within range.
3. $$ 16 - 3k + 5 = 0 $$ → $$ 21 - 3k = 0 $$ → $$ 3k = 21 $$ → $$ k = 7 $$, which is an integer and within range.

So, only cases 2 and 3 give valid $$ k $$ values. Now compute their contributions.

Case 2: $$ k = 5 $$, multiplied by $$ -\frac{1}{x} $$

First, find $$ T_5 $$:

$$ T_5 = \binom{8}{5} 2^{8-5} (-1)^5 x^{16 - 3 \times 5} = \binom{8}{5} 2^{3} (-1)^5 x^{16-15} = \binom{8}{5} \cdot 8 \cdot (-1) x^{1} $$

Calculate $$ \binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 $$, so:

$$ T_5 = 56 \cdot 8 \cdot (-1) \cdot x = -448x $$

Now multiply by $$ -\frac{1}{x} $$:

$$ \left(-\frac{1}{x}\right) \times (-448x) = (-1) \times (-448) \times \frac{x}{x} = 448 \times 1 = 448 $$

Contribution is 448.

Case 3: $$ k = 7 $$, multiplied by $$ 3x^5 $$

First, find $$ T_7 $$:

$$ T_7 = \binom{8}{7} 2^{8-7} (-1)^7 x^{16 - 3 \times 7} = \binom{8}{7} 2^{1} (-1)^7 x^{16-21} = \binom{8}{7} \cdot 2 \cdot (-1) x^{-5} $$

Calculate $$ \binom{8}{7} = 8 $$, so:

$$ T_7 = 8 \cdot 2 \cdot (-1) \cdot x^{-5} = -16x^{-5} $$

Now multiply by $$ 3x^5 $$:

$$ (3x^5) \times (-16x^{-5}) = 3 \times (-16) \times x^{5 + (-5)} = -48 \times x^0 = -48 $$

Contribution is -48.

Sum the contributions: $$ 448 + (-48) = 400 $$.

Hence, the term independent of $$ x $$ is 400.

Comparing with the options:

A. $$-496$$

B. $$-400$$

C. $$496$$

D. $$400$$

So, the correct answer is Option D.