Pyramid cube melting

They melted the steel part in the shape of a truncated quadrilateral needle and produced three identical cubes. Determine the surface area of one cube if the edges of the bases of the pyramid are 30 mm and 80 mm and the pyramid's height is 60 mm. I don't know how to deal with it, and I haven't found a solution anywhere, so I'm turning to you.

Final Answer:

S =  9666.5515 mm²

Step-by-step explanation:

$$\begin{gathered} a=30\text{mm} \\ b=80\text{mm} \\ h=60\text{mm} \\ {S}_1=a^2=30^2=900{\text{mm}}^2 \\ {S}_2=b^2=80^2=6400{\text{mm}}^2 \\ V=\frac{h}{3}\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)=\frac{60}{3}\cdot (900+\sqrt{900\cdot 6400}+6400)=194000{\text{mm}}^3 \\ V=3V_1 \\ {V}_1=V/3=194000/3=\frac{194000}{3}\doteq 64666.6667{\text{mm}}^3 \\ {V}_1=x^3\text{mm} \\ x=\sqrt[3]{V_1}=\sqrt[3]{64666.6667}\doteq 40.1384\text{mm} \\ S=6\cdot x^2=6\cdot 40.1384^2=9666.5515{\text{mm}}^2 \end{gathered}$$

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