Since $x$ is inversely proportional to the cube of $y$, we can write this relationship as: $x \propto \frac{1}{y^3}$ This can be expressed with a constant $k$: $x = \frac{k}{y^3}$

Given that $x = 9$ for a particular value of $y$, let's denote this specific value of $y$ as $y_1$. Therefore: $9 = \frac{k}{y_1^3}$

From this, we can solve for $k$: $k = 9 y_1^3$

Now, we are asked to find the value of $x$ when $y$ is increased by 100%. If $y$ is increased by 100%, the new value of $y$ becomes: $y_2 = 2y_1$

Using the relationship $x = \frac{k}{y^3}$ with the new value of $y$: $x_2 = \frac{k}{(2y_1)^3}$ $x_2 = \frac{9 y_1^3}{8 y_1^3}$ $x_2 = \frac{9}{8}$

Thus, the new value of $x$ is $\frac{9}{8}$.

Next, we calculate the percentage decrease in the value of $x$ from the original value (9) to the new value ($\frac{9}{8}$):

1. Find the difference in values: $\Delta x = 9 - \frac{9}{8}$ $\Delta x = 9 - 1.125$ $\Delta x = 7.875$

2. Calculate the percentage decrease: $\text{Percentage decrease} = \left( \frac{\Delta x}{\text{original } x} \right) \times 100\%$ $\text{Percentage decrease} = \left( \frac{7.875}{9} \right) \times 100\%$ $\text{Percentage decrease} \approx 87.5\%$

So, when $y$ is increased by 100%, the value of $x$ decreases by approximately 87.5%.

Would you like more details or have any questions?

Here are some related questions you might be interested in:

1. How do you solve problems involving direct proportionality?
2. Can you explain how to solve problems involving joint variation?
3. What are some real-life applications of inverse proportionality?
4. How do you derive the constant of proportionality in different scenarios?
5. What is the difference between linear and non-linear proportional relationships?

Tip: When dealing with proportional relationships, always identify the type of proportionality (direct or inverse) and the variables involved to set up the correct equation.