### Instruction: Suppose that $a^2$ varies inversely with $b^3$. If $a=7$ when $b=3$, find the value of $a^2$ when $b=6$.### Response: I know that inverse variation means that the product of the two quantities is constant, so I can write an equation of the form $a^2k=b^3$, where $k$ is some constant.
To find $k$, I can plug in the given values of $a$ and $b$: $7^2k=3^3$, which simplifies to $49k=27$.
Solving for $k$, I get $k=\frac{27}{49}$.
Now, I can use this value of $k$ and the given value of $b=6$ to find the corresponding value of $a^2$: $a^2\frac{27}{49}=6^3$, which simplifies to $a^2=\frac{49}{27}6^3$.
Evaluating the right-hand side, I get $a^2=392$.