Expand (x-2y)³ using the binomial theorem.

Step 1: Expand $(x-2y)^3$ using the binomial theorem.
The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
For $(x-2y)^3$, we have $a=x$, $b=-2y$, and $n=3$.

$$(x-2y)^3 = \binom{3}{0}x^3(-2y)^0 + \binom{3}{1}x^2(-2y)^1 + \binom{3}{2}x^1(-2y)^2 + \binom{3}{3}x^0(-2y)^3$$ $$= (1)x^3(1) + (3)x^2(-2y) + (3)x(4y^2) + (1)(1)(-8y^3)$$ $$= x^3 - 6x^2y + 12xy^2 - 8y^3$$

Step 2: Multiply the expanded form by $x^2$.

$$x^2(x-2y)^3 = x^2(x^3 - 6x^2y + 12xy^2 - 8y^3)$$ $$= x^2 \cdot x^3 - x^2 \cdot 6x^2y + x^2 \cdot 12xy^2 - x^2 \cdot 8y^3$$ $$= x^5 - 6x^4y + 12x^3y^2 - 8x^2y^3$$

The binomial expansion is:

$$\boxed{x^5 - 6x^4y + 12x^3y^2 - 8x^2y^3}$$

Step 1: Write the general term for the expansion of $\left(x - \frac{3}{x}\right)^4$.
The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
Here, $a=x$, $b=-\frac{3}{x} = -3x^{-1}$, and $n=4$.

$$T_{r+1} = \binom{4}{r} (x)^{4-r} (-3x^{-1})^r$$ $$T_{r+1} = \binom{4}{r} x^{4-r} (-3)^r (x^{-1})^r$$ $$T_{r+1} = \binom{4}{r} (-3)^r x^{4-r-r}$$ $$T_{r+1} = \binom{4}{r} (-3)^r x^{4-2r}$$

Step 2: Find the value of $r$ for the term independent of $x$.
For the term independent of $x$, the power of $x$ must be 0.

$$4-2r = 0$$ $$2r = 4$$ $$r = 2$$

Step 3: Substitute $r=2$ into the general term formula.

$$T_{2+1} = T_3 = \binom{4}{2} (-3)^2 x^{4-2(2)}$$ $$T_3 = \frac{4!}{2!(4-2)!} (9) x^0$$ $$T_3 = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} \times 9 \times 1$$ $$T_3 = \frac{24}{4} \times 9$$ $$T_3 = 6 \times 9$$ $$T_3 = 54$$

The term independent of $x$ is:

$$\boxed{54}$$

Step 1: Use the binomial series expansion formula for $(1+x)^n$.
The binomial series expansion is given by $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$.
For $(1-y)^{\frac{1}{3}}$, we have $x=-y$ and $n=\frac{1}{3}$.

Step 2: Calculate the first term.
The first term is 1.

Step 3: Calculate the second term.

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