The statement says that $y$ varies jointly as $x$ and the cube of $z$. The general form for this type of variation is:

$y = kxz^3$

where $k$ is the constant of variation.

We are given that:

• $y = 405$
• $x = 3$
• $z = 3$

Substitute these values into the equation to solve for $k$:

$405 = k(3)(3^3)$

Since $3^3 = 27$, the equation becomes:

$405 = k(3)(27)$

$405 = 81k$

Now, solve for $k$:

$k = \frac{405}{81} = 5$

Thus, the constant of variation $k$ is:

$k = 5$

Part 2: The variation equation

Now that we know $k = 5$, substitute it back into the variation equation:

$y = 5xz^3$

So the variation equation is:

$y = 5xz^3$

Would you like further details or have any questions about this?

Here are 5 related questions to practice:

1. If $y$ varies jointly as $x$ and the square of $z$, how would the equation change?
2. What happens to $y$ if $z$ is doubled while $x$ stays constant?
3. Solve for $y$ when $x = 2$, $z = 4$, and $k = 5$.
4. How do you identify when a problem involves joint variation instead of direct or inverse variation?
5. Can you rewrite the equation if $y$ varies jointly as $x$, $z^3$, and inversely as $w$?

Tip: Joint variation problems involve multiple variables, where one variable depends on the product (or quotient) of others.