This problem asks to find the expression $(h + p)(-3)$, where:

• $h(x) = x^2$ and $D = \text{Reals}$,
• $p(x) = x^3$ and $D = \text{Negative integers}$.

Let’s break it down:

1. $(h + p)(x) = h(x) + p(x)$, so we have: $(h + p)(x) = x^2 + x^3.$

2. Now, we want to evaluate this at $x = -3$: $(h + p)(-3) = (-3)^2 + (-3)^3.$

3. Calculate each part: $(-3)^2 = 9 \quad \text{and} \quad (-3)^3 = -27.$

4. So, the result is: $(h + p)(-3) = 9 + (-27) = -18.$

Thus, the value of $(h + p)(-3)$ is $-18$.

Do you need any further clarification or details on the steps?

Here are five related questions:

1. What happens when you evaluate $(h + p)(2)$?
2. How would you solve a similar problem where $p(x) = x^4$?
3. What is the general form for the sum of two functions evaluated at a point?
4. How do the domains of the functions affect the problem?
5. What are the different ways to combine functions, such as addition and multiplication?

Tip: When evaluating sums or differences of functions, always be careful to apply the operations correctly to each function before evaluating.