Jointly and cube power

If y varies jointly as x and the cube of z and y=16 when x=4 and z=2, find an equation representing this relationship.
a. y=xz
b. y = 2xz³
c.y = 2kx/z³
d.y = xz³/2​

Correct answer:

y = x z^3

Step-by-step explanation:

$$\begin{gathered} y_1=16 \\ {x}_1=4 \\ {z}_1=2 \\ y=k\cdot x\cdot z^3 \\ k=\frac{y_1}{x_1\cdot z_1^2}=\frac{16}{4\cdot 2^2}=1 \\ y=k\cdot x\cdot z^3 \\ y=x{z}^3 \end{gathered}$$

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