Solve $abc=2(a-2)(b-2)(c-2)$ where $a,b $ and $c$ are integers
I am trying to find the integers $a,b$ and $c$ so that the number of hidden cubes in a rectangular cube with dimensions $a,b,c$ (built from $1\times 1$ cubes) to be half of the total number of cubes. The number of hidden cubes are $$(a-2)(b-2)(c-2)$$
So we must have $$(a-2)(b-2)(c-2)=\frac{abc}{2} \Rightarrow \ \ 2(a−2)(b−2)(c−2)= abc $$
But I don't know how to solve this equation for $a,b$ and $c$.
Erica89
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Erdos
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The answer is accepted.
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