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Let's work through the two problems step by step.

### Question 8:
- **Problem**: \( y \) is inversely proportional to the cube of \( x \).
  - This means \( y \propto \frac{1}{x^3} \), or \( y = \frac{k}{x^3} \), where \( k \) is the constant of proportionality.
  
- Given: \( y = 250 \) when \( x = 0.2 \).
  
  We can use this to find \( k \):
  \[
  250 = \frac{k}{(0.2)^3}
  \]
  Solving for \( k \):
  \[
  250 = \frac{k}{0.008} \quad \Rightarrow \quad k = 250 \times 0.008 = 2
  \]
  
- Now, using \( k = 2 \), we can find the value of \( y \) when \( x = 0.5 \):
  \[
  y = \frac{2}{(0.5)^3} = \frac{2}{0.125} = 16
  \]

Thus, the value of \( y \) when \( x = 0.5 \) is **16**.

---

### Question 9:
- **Problem**: \( x \) is directly proportional to the cube of \( y \).
  - This means \( x \propto y^3 \), or \( x = k y^3 \), where \( k \) is the constant of proportionality.
  
- Given: \( x = 32 \) when \( y = 0.4 \).
  
  We can use this to find \( k \):
  \[
  32 = k (0.4)^3
  \]
  Solving for \( k \):
  \[
  32 = k \times 0.064 \quad \Rightarrow \quad k = \frac{32}{0.064} = 500
  \]
  
- Now, using \( k = 500 \), we can find the value of \( y \) when \( x = 256 \):
  \[
  256 = 500 y^3 \quad \Rightarrow \quad y^3 = \frac{256}{500} = 0.512
  \]
  Taking the cube root of both sides:
  \[
  y = \sqrt[3]{0.512} = 0.8
  \]

Thus, the value of \( y \) when \( x = 256 \) is **0.8**.

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Would you like further details on any step or concept? Here are some related questions:

1. How do you derive proportionality relationships?
2. Can we explore more about inverse and direct proportionality?
3. What are some common mistakes to avoid when working with proportionality problems?
4. How would the results change if the exponents were different?
5. How can cube roots be calculated without a calculator?

**Tip**: Always double-check units and signs in proportionality equations to ensure accuracy!

8. y is inversely proportional to the cube of x. When y = 250, x = 0.2, find the value of y when x = 0.5. 9. x is directly proportional to the cube of y. When x = 32, y = 0.4, find the value of y when x = 256.

Learn how to solve problems involving direct and inverse proportionality with cubic relationships. This includes working through detailed solutions where y is inversely proportional to the cube of x, and x is directly proportional to the cube of y. Suitable for Grades 9-11.

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