Question 232·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

The negative exponent on $z$ means a reciprocal, not a negative value. Because $z$ is nonzero, $z^{-2}=\frac{1}{z^2}$, so the given equation becomes:

To remove the $z^2$ beneath $5k^3$, multiply both sides by $z^2$:

The $5$ multiplies $k^3$, so divide both sides by $5$:

Pick $y=5$ and $z=2$ as nonzero test values; $z=2$ makes its square matter. Type $y_1=5$, $z=2$, and $y_1\sim5k_1^3z^{-2}$. The $\sim$ tells Desmos to solve the given equation for $k_1$. Under PARAMETERS, it shows $k_1\approx1.5874$.

Add $k_1^3$ and $\frac{y_1z^2}{5}$ to compare the two sides of your rearranged equation. Desmos prints $4$ for each. The sample catches a possible algebra slip, but a match at one pair of values doesn't prove a formula.

Take the cube root of both sides of $\frac{yz^2}{5}=k^3$:

Cubing preserves a number's sign, so there is one real cube root, not a plus-or-minus pair. This expresses $k$ in terms of $y$ and $z$. Choice C.