The relationship between $y$ and the cube of $x$ is a direct variation, which can be written as:

$y = kx^3$

Where $k$ is the constant of proportionality. To find $k$, we use the given values $y = 25$ when $x = 2$:

$25 = k \cdot 2^3$ $25 = k \cdot 8$ $k = \frac{25}{8}$

Now, to find $y$ when $x = 6$, substitute $x = 6$ into the equation:

$y = \frac{25}{8} \cdot 6^3$ $y = \frac{25}{8} \cdot 216$ $y = \frac{25 \cdot 216}{8}$ $y = 25 \cdot 27$ $y = 675$

Thus, $y = 675$ when $x = 6$.

Would you like further details or have any questions?

Here are 5 related questions for practice:

1. If $y$ varies directly with $x^4$, and $y = 32$ when $x = 2$, find $y$ when $x = 4$.
2. Given $y$ varies inversely with $x$, and $y = 10$ when $x = 5$, find $y$ when $x = 2$.
3. If $y$ varies directly with the square of $x$, and $y = 9$ when $x = 3$, find $y$ when $x = 5$.
4. Solve for $y$ when $y$ varies directly with $x^2$, given $y = 50$ and $x = 5$, and $x = 10$.
5. If $y$ varies inversely with the square of $x$, and $y = 16$ when $x = 2$, find $y$ when $x = 4$.

Tip: Always solve for the constant of variation first before solving for other values of $x$.