AIME I 2000 Problem 8: Calculating the height of the liquid given Fraction of the Volume

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As the title says, I'm looking at problem number 8 from AIME I 2000.

https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_8

I'm currently looking at solution 3 and I understand everything up to the point where it says the cone that is filled with air must have a height of the cube root of the fraction of the cone that the air occupies.

What am I missing here? Where does one get cube root from anything?

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For a given right circular cone, the ratio of the radius $r$ to the height $h$ is constant. In this instance that ratio is $\frac{r}{h} = \frac{5}{12} = k$. The volume $V_1$ at a given height $h_1$ is thus $V_1 = \frac{\pi (kh_1)^2 h_1}{3}=\frac{\pi k^2}{3} h_1^3$. The volume at another height $h_2$ is likewise $V_2=\frac{\pi k^2}{3} h_2^3$. If the ratio of volumes is known, i.e. $\frac{V_1}{V_2} = C$, the ratio of the heights will also be known, i.e. $h_1 = C^{\frac{1}{3}}h_2$.