20(0.25)=[(16sqrt(2)/3)-(sqrt(3)/16)(4sqrt(2/3)-h)^3]

It seems like you're using a different approach to solve the problem, which involves equating the water volume at a certain time to an expression involving the total volume of the tetrahedral tank minus the volume of a smaller tetrahedron formed by the unfilled portion. Let's break down the equation:

### Given Equation:
\[
20(0.25) = \frac{16\sqrt{2}}{3} - \frac{\sqrt{3}}{16}\left(4\sqrt{\frac{2}{3}} - h\right)^3
\]

Where:
- \( 20(0.25) \) represents the volume of water added after 20 minutes.
- \( \frac{16\sqrt{2}}{3} \) is the total volume of the tetrahedral tank.
- \( \frac{\sqrt{3}}{16}\left(4\sqrt{\frac{2}{3}} - h\right)^3 \) represents the volume of the unfilled portion of the tank.

### Step 1: Simplify the Equation
Let's first simplify the left side:
\[
20(0.25) = 5 \text{ cubic meters}
\]

So the equation becomes:
\[
5 = \frac{16\sqrt{2}}{3} - \frac{\sqrt{3}}{16}\left(4\sqrt{\frac{2}{3}} - h\right)^3
\]

### Step 2: Isolate the Unfilled Volume
Rearrange the equation to isolate the unfilled volume term:
\[
\frac{\sqrt{3}}{16}\left(4\sqrt{\frac{2}{3}} - h\right)^3 = \frac{16\sqrt{2}}{3} - 5
\]

Calculate the right side:
\[
\frac{16\sqrt{2}}{3} - 5 = \frac{16\sqrt{2}}{3} - \frac{15}{3} = \frac{16\sqrt{2} - 15}{3}
\]

So the equation becomes:
\[
\frac{\sqrt{3}}{16}\left(4\sqrt{\frac{2}{3}} - h\right)^3 = \frac{16\sqrt{2} - 15}{3}
\]

### Step 3: Solve for \( h \)
Multiply both sides by 16 and divide by \( \sqrt{3} \):
\[
\left(4\sqrt{\frac{2}{3}} - h\right)^3 = \frac{16 \times 16\sqrt{2} - 15 \times 16}{3\sqrt{3}} 
\]

Simplify the right-hand side, and then take the cube root to solve for \( h \).

Would you like to proceed with the calculation, or do you want any further details on any part of this process?

Explore the solution process for solving a volume equation involving cube roots step-by-step. This problem involves equations related to the volume of a tetrahedral tank and requires understanding cube roots and volume calculations. Suitable for advanced high school students and above.

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