Answer :

To make [tex]\( h \)[/tex] the subject of the given relation:
[tex]\[ V = \pi h^2 \frac{(r - h)}{3} \][/tex]

Follow these steps:

1. Multiply both sides by 3 to eliminate the denominator:
[tex]\[ 3V = \pi h^2 (r - h) \][/tex]

2. Divide both sides by [tex]\( \pi \)[/tex]:
[tex]\[ \frac{3V}{\pi} = h^2 (r - h) \][/tex]

3. Expand the right-hand side:
[tex]\[ \frac{3V}{\pi} = h^2 r - h^3 \][/tex]

4. Rearrange into a standard cubic equation form:
[tex]\[ h^3 - h^2 r + \frac{3V}{\pi} = 0 \][/tex]

Now, the above cubic equation:
[tex]\[ h^3 - h^2 r + \frac{3V}{\pi} = 0 \][/tex]
is what we need to solve for [tex]\( h \)[/tex].

The solutions to this cubic equation for [tex]\( h \)[/tex] are:
[tex]\[ h = -\frac{r^2}{3 \left( \sqrt[3]{81V / (2 \pi) - r^3 + \sqrt{ -4r^6 + \left( 81V/\pi - 2r^3 \right)^2} / 2} \right)} + \frac{r}{3} - \frac{\sqrt[3]{81V / (2\pi) - r^3 + \sqrt{ -4r^6 + \left( 81V/\pi - 2r^3 \right)^2 } / 2}}{3} \][/tex]
and the two complex conjugate solutions:
[tex]\[ h = -\frac{r^2}{3 \left( -\frac{1}{2} - \frac{\sqrt{3}i}{2} \right) \sqrt[3]{81V / (2\pi) - r^3 + \sqrt{ -4r^6 + ( 81V/\pi - 2r^3 )^2 } / 2}} + \frac{r}{3} - \left( -\frac{1}{2} - \frac{\sqrt{3}i}{2} \right) \frac{\sqrt[3]{81V / (2\pi) - r^3 + \sqrt{ -4r^6 + ( 81V/\pi - 2r^3 )^2 } / 2}}{3} \][/tex]
and
[tex]\[ h = -\frac{r^2}{3 \left( -\frac{1}{2} + \frac{\sqrt{3}i}{2} \right) \sqrt[3]{81V / (2\pi) - r^3 + \sqrt{ -4r^6 + ( 81V/\pi - 2r^3 )^2 } / 2}} + \frac{r}{3} - \left( -\frac{1}{2} + \frac{\sqrt{3}i}{2} \right) \frac{\sqrt[3]{81V / (2 \pi) - r^3 + \sqrt{ -4r^6 + ( 81V/\pi - 2r^3 )^2 } / 2}}{3} \][/tex]