Independent Term

General term in the expansion is

$$T_{r+1}=\binom{12}{r}x^{12-r}\left(\frac3{x^2}\right)^r$$

$$=\binom{12}{r}3^r x^{12-3r}$$

Coefficient of

$$x^9$$

occurs when

$$12-3r=9$$

$$3r=3$$

$$r=1$$

Hence corresponding term is

$$T_2=\binom{12}{1}3^1x^9$$

Therefore coefficient of

$$x^9$$

is

$$12\times3=36$$

Now term independent of

$$x$$

occurs when

$$12-3r=0$$

$$r=4$$

Hence constant term is

$$T_5=\binom{12}{4}3^4$$

$$=495\times81$$

Therefore required ratio is

$$36:495\times81$$

Divide by

$$9$$

$$=4:4455$$

Hence, the required ratio is

$$\boxed{4:4455}$$

To find the term independent of $$x$$ in the expansion, we first simplify the two fractions inside the parentheses.

Step 1: Simplify the First Fraction

Let the first term be $$A$$:

$$A = \frac{x+1}{x^{2/3} - x^{1/3} + 1}$$

Using the sum of cubes factorization $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$, we can write $$x + 1$$ as:

$$x + 1 = (x^{1/3})^3 + 1^3 = (x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)$$

Substituting this back into $$A$$:

$$A = \frac{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$$

Step 2: Simplify the Second Fraction

Let the second term be $$B$$:

$$B = \frac{x-1}{x - x^{1/2}}$$

Using the difference of squares $$a^2 - b^2 = (a - b)(a + b)$$ and factoring the denominator:

• Numerator: $$x - 1 = (x^{1/2} - 1)(x^{1/2} + 1)$$
• Denominator: $$x - x^{1/2} = x^{1/2}(x^{1/2} - 1)$$

Substituting these back into $$B$$:

$$B = \frac{(x^{1/2} - 1)(x^{1/2} + 1)}{x^{1/2}(x^{1/2} - 1)} = \frac{x^{1/2} + 1}{x^{1/2}} = 1 + x^{-1/2}$$

Step 3: Set up the Binomial Expansion

Now, we subtract the simplified terms:

$$A - B = (x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$$

We need to find the term independent of $x$ in the expansion of:

$$(x^{1/3} - x^{-1/2})^{10}$$

The general term $$T_{r+1}$$ is given by:

$$T_{r+1} = \binom{10}{r} \left(x^{1/3}\right)^{10-r} \left(-x^{-1/2}\right)^r$$  $$T_{r+1} = \binom{10}{r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$$

Step 4: Find $$r$$ and Compute the Term

For the term to be independent of $$x$$, the exponent of $$x$$ must be $$0$$:

$$\frac{10-r}{3} - \frac{r}{2} = 0$$

Multiply the entire equation by $$6$$ to clear the denominators:

$$2(10-r) - 3r = 0$$ $$20 - 2r - 3r = 0$$ $$5r = 20 \implies r = 4$$

Now substitute $$r = 4$$ back into the general term formula:

$$T_{5} = \binom{10}{4} (-1)^4 = \binom{10}{4}$$ $$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$

Final Answer

The term independent of $$x$$ is $$210$$.

The general term $$T_{r+1}$$ in the binomial expansion of $$(a + b)^n$$ is given by:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

For our expression $$\left(2x^2 + \frac{1}{x}\right)^{15}$$, we identify:

• $$a = 2x^2$$
• $$b = \frac{1}{x} = x^{-1}$$
• $$n = 15$$

Substitute these into the formula:

$$T_{r+1} = \binom{15}{r} (2x^2)^{15-r} \left(x^{-1}\right)^r$$

Expand the terms to group all exponents of $$x$$ together:

$$T_{r+1} = \binom{15}{r} \cdot 2^{15-r} \cdot \left(x^2\right)^{15-r} \cdot x^{-r}$$ $$T_{r+1} = \binom{15}{r} \cdot 2^{15-r} \cdot x^{2(15-r)} \cdot x^{-r}$$ $$T_{r+1} = \binom{15}{r} \cdot 2^{15-r} \cdot x^{30 - 2r - r}$$ $$T_{r+1} = \binom{15}{r} \cdot 2^{15-r} \cdot x^{30 - 3r}$$

For a term to be independent of $$x$$, the total exponent of $$x$$ must be equal to $$0$$ ($$x^0 = 1$$).

Set the power of $$x$$ to $$0$$:

$$30 - 3r = 0$$ $$30 = 3r$$ $$r = 10$$

Conclusion

Since $$r = 10$$, the term independent of $$x$$ is the $$11^{\text{th}}$$ term ($$T_{10+1}$$). The question asks for the value of $$r$$, which is exactly 10.

This matches Option C.