Find the binomial expansion of x³(x-2y)⁴.

1. Find the binomial expansion of $x^3(x-2y)^4$.

• Step 1: Expand $(x-2y)^4$ using the binomial theorem. The binomial theorem states that $(A+B)^n = \sum_{r=0}^{n} \binom{n}{r} A^{n-r} B^r$. Here, $A=x$, $B=-2y$, and $n=4$.

  $$(x-2y)^4 = \binom{4}{0}x^4(-2y)^0 + \binom{4}{1}x^3(-2y)^1 + \binom{4}{2}x^2(-2y)^2 + \binom{4}{3}x^1(-2y)^3 + \binom{4}{4}x^0(-2y)^4$$ $$= 1 \cdot x^4 \cdot 1 + 4 \cdot x^3 \cdot (-2y) + 6 \cdot x^2 \cdot (4y^2) + 4 \cdot x \cdot (-8y^3) + 1 \cdot 1 \cdot (16y^4)$$ $$= x^4 - 8x^3y + 24x^2y^2 - 32xy^3 + 16y^4$$

• Step 2: Multiply the expansion by $x^3$.

  $$x^3(x^4 - 8x^3y + 24x^2y^2 - 32xy^3 + 16y^4)$$ $$= x^3 \cdot x^4 - x^3 \cdot 8x^3y + x^3 \cdot 24x^2y^2 - x^3 \cdot 32xy^3 + x^3 \cdot 16y^4$$ $$= x^7 - 8x^6y + 24x^5y^2 - 32x^4y^3 + 16x^3y^4$$

  The binomial expansion is $\boxed{x^7 - 8x^6y + 24x^5y^2 - 32x^4y^3 + 16x^3y^4}$.

2. Find the term independent of $x$ in the expansion of $(x-\frac{3}{x})^4$.

• Step 1: Write the general term of the expansion. 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 to be 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 to find the term.

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

  The term independent of $x$ is $\boxed{54}$.

3. Find the first three terms in ascending powers of $y$ in the expansion of $(1-y)^{\frac{1}{3}}$.

• Step 1: Use the generalized binomial theorem. The generalized binomial theorem for $(1+x)^n$ is $1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$. Here, $x=-y$ and $n=\frac{1}{3}$.

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

• Step 3: Calculate the second term.

  $$nx = \left(\frac{1}{3}\right)(-y) = -\frac{1}{3}y$$

• Step 4: Calculate the third term.

  $$\frac{n(n-1)}{2!}x^2 = \frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2 \cdot 1}(-y)^2$$ $$= \frac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2} y^2$$ $$= \frac{-\frac{2}{9}}{2} y^2$$ $$= -\frac{1}{9}y^2$$

• Step 5: Combine the first three terms. The first three terms in ascending powers of $y$ are $\boxed{1 - \frac{1}{3}y - \frac{1}{9}y^2}$.

4. Find the coefficient of $x^{-3}$ in the expansion of $(x+1)(2x-\frac{1}{x})^3$.

• Step 1: Expand $(2x-\frac{1}{x})^3$. Let $A=2x$ and $B=-\frac{1}{x} = -x^{-1}$. Using $(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$: $$(2x-x^{-1})^3 = (2x)^3 + 3(2x)^2(-x^{-1}) + 3(2x)(-x^{-1})^2 + (-x^{-1})^3$$ $$= 8x^3 + 3(4x^2)(-x^{-1}) + 3(2x)(x^{-2}) - x^{-3}$$ $$$$