A geometry problem by Dipa Parameswara

Let the length of the cuboid be $x \space \text{cm}$ , the width of the cuboid be $y \space \text{cm}$ and the height of the cuboid be $z \space \text{cm}$ .

The volume of the cuboid is $V = xyz = 288 \space \text{cm}^3 \space ...(1)$

The surface are is $A = 2xy + 2yz + 2xz = 288 \space \text{cm}^2 \rightarrow xy + yz + xz = 144 \space ...(2)$

$(2):(1)$ gives $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{2}$

One of the algebraic identities for the Egyptian fraction decomposition is $\dfrac{1}{a} = \dfrac{1}{a(a+1)} + \dfrac{1}{a+1}$ .

By this identity, we can find that $(x,y,z) = (4,6,12)$ . So, The answer is $\boxed{\boxed{22}} \space \text{cm}$ .

The sides is 4,6,12 so the sum is 22