I see you've uploaded an image with several math problems. I'll go ahead and solve them one by one starting with question 8a.

8a. Expand $\frac{1}{(1 + 2x)^2}$ up to the term in $x^3$.

This is a binomial expansion problem. We can rewrite $\frac{1}{(1 + 2x)^2}$ as $(1 + 2x)^{-2}$, then use the binomial series expansion formula: $(1 + u)^n = 1 + n u + \frac{n(n-1)}{2} u^2 + \frac{n(n-1)(n-2)}{6} u^3 + \cdots$ where $u = 2x$ and $n = -2$.

Now we expand: $(1 + 2x)^{-2} = 1 + (-2)(2x) + \frac{(-2)(-3)}{2}(2x)^2 + \frac{(-2)(-3)(-4)}{6}(2x)^3 + \cdots$

Simplifying the terms: $= 1 - 4x + 12x^2 - 32x^3 + \cdots$

Thus, the expansion up to the term in $x^3$ is: $\frac{1}{(1 + 2x)^2} = 1 - 4x + 12x^2 - 32x^3$

8b. Find the coefficient of $x^2$ in the expansion of $\frac{(1 - 3x)^3}{(1 + 2x)^2}$.

We already have the expansion for $\frac{1}{(1 + 2x)^2}$, which is $1 - 4x + 12x^2 - 32x^3$.

Now we expand $(1 - 3x)^3$ up to the term in $x^2$: $(1 - 3x)^3 = 1 + 3(-3x) + \frac{3(3-1)}{2}(-3x)^2 = 1 - 9x + 27x^2$

To find the expansion of $\frac{(1 - 3x)^3}{(1 + 2x)^2}$, we multiply the expansions of $(1 - 3x)^3$ and $\frac{1}{(1 + 2x)^2}$ up to the term in $x^2$: $(1 - 9x + 27x^2)(1 - 4x + 12x^2) = 1 - 9x + 27x^2 - 4x + 36x^2 - 108x^3 + \cdots$

Combining like terms: $= 1 - 13x + 63x^2 + \cdots$

Thus, the coefficient of $x^2$ is 63.

Would you like to continue with the next part or have any further questions on this? Also, here are some related practice questions you might find helpful:

1. Expand $\frac{1}{(1 - x)^3}$ up to $x^3$.
2. Find the coefficient of $x^2$ in the expansion of $(1 + 4x)^3 (1 - 2x)^2$.
3. What is the general binomial expansion of $(1 + ax)^n$?
4. Estimate $\frac{1}{(1 + 0.1)^2}$ using a binomial expansion.
5. Determine for what values of $x$ the expansion $(1 - 2x)^{-2}$ converges.

Tip: When expanding binomials, pay attention to negative powers and coefficients for accurate results.