Two cubes, each of volume $729 c m^{3}$ are joined end to end. Find the surface areas of the resulting cuboid.

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Solution

The volume of cube $729$
The side of cube is $\sqrt[3]{729} = 9 c m$
The length of resulting cuboid $9 + 9 = 18 c m$
The surface area is $2 (9 (9) + 9 (18) + 9 (18)) 2 (81 + 162 + 162) 2 (405) = 810 c m^{2}$

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