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The expression in the problem asks to compute \( a^3b \) where \( a = -2i \) and \( b = 5 - 5i \).

Let's go step by step:

1. **Step 1: Compute \( a^3 \)**  
   First, we need to calculate \( a^3 \), where \( a = -2i \).

   \[
   a = -2i \quad \Rightarrow \quad a^3 = (-2i)^3
   \]
   Now use the fact that \( i^2 = -1 \) and \( i^3 = -i \):

   \[
   (-2i)^3 = (-2)^3 \cdot i^3 = -8 \cdot (-i) = 8i
   \]

   So, \( a^3 = 8i \).

2. **Step 2: Multiply \( a^3 \) by \( b \)**  
   Now, we multiply \( a^3 = 8i \) by \( b = 5 - 5i \):

   \[
   a^3b = 8i(5 - 5i)
   \]

   Use the distributive property to expand this:

   \[
   a^3b = 8i \cdot 5 + 8i \cdot (-5i)
   \]
   \[
   a^3b = 40i + 40
   \]
   (Since \( i^2 = -1 \), the second term becomes \( 40 \), not \( -40 \)).

3. **Final Answer**:
   \[
   a^3b = 40 + 40i
   \]

Thus, the fully simplified form of \( a^3b \) is \( \boxed{40 + 40i} \).

Would you like further explanation or clarification on any part of the solution? Here are five related questions you might find interesting:

1. How does the imaginary unit \( i \) behave in higher powers, like \( i^4 \), \( i^5 \), etc.?
2. What happens if we change the values of \( a \) and \( b \)? How does that affect \( a^3b \)?
3. How do complex numbers like \( 5 - 5i \) relate to their geometric representation on the complex plane?
4. How would you calculate the magnitude of the complex number result \( 40 + 40i \)?
5. Can you extend the calculation to find \( a^4b \) or \( a^5b \) for practice?

**Tip**: Always remember that powers of the imaginary unit \( i \) follow a cyclic pattern: \( i, -1, -i, 1 \), repeating every four steps!

If a = -2i and b = 5 - 5i, then find the value of the a^3b in fully simplified form.

Find the value of a^3b where a = -2i and b = 5 - 5i, using complex number multiplication and powers of i. This solution involves step-by-step simplification for students in Grades 10-12.

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