To solve this problem, we can use the formula for direct variation. Since $y$ varies directly as the cube of $x$, we can express this relationship as:

$y = kx^3$

where $k$ is the constant of proportionality.

Step 1: Find the constant $k$.

We know that when $x = 2$, $y = 5$. Substituting these values into the equation:

$5 = k(2^3)$

$5 = k(8)$

$k = \frac{5}{8}$

Step 2: Find $y$ when $x = 3$.

Now that we know $k = \frac{5}{8}$, we can find $y$ when $x = 3$ by substituting $x = 3$ into the equation:

$y = \frac{5}{8}(3^3)$

$y = \frac{5}{8}(27)$

$y = \frac{135}{8}$

$y = 16.875$

Thus, when $x = 3$, $y$ is 16.875.

Would you like more details or have any questions about this problem?

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Here are some related questions for practice:

1. If $y$ varies directly as the square of $x$, and $y = 12$ when $x = 4$, find $y$ when $x = 5$.
2. If $y$ varies directly as $x$ and $y = 20$ when $x = 4$, find $y$ when $x = 10$.
3. If $y$ varies directly as the fourth power of $x$ and $y = 81$ when $x = 3$, find $y$ when $x = 2$.
4. Suppose $y$ varies inversely as $x$, and $y = 10$ when $x = 2$. Find $y$ when $x = 5$.
5. If $y$ varies directly as $x^2$ and $y = 25$ when $x = 5$, find $x$ when $y = 100$.
6. If $y$ varies directly as the square root of $x$, and $y = 7$ when $x = 49$, find $y$ when $x = 16$.
7. If $y$ varies directly as $x^3$ and $y = 64$ when $x = 4$, find $y$ when $x = 2$.
8. Suppose $y$ varies directly as $x^n$ and $y = 32$ when $x = 2$ for a certain $n$. Find the value of $n$ if $y = 512$ when $x = 4$.

Tip: In direct variation problems, finding the constant of proportionality first can simplify solving the problem.