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 $c$.
16
Answer
Answers can only be viewed under the following conditions:
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.

4.8K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 1164 views
- $5.00
Related Questions
- Attempting to make a formula/algorithm based on weighted averages to find how much equipment we need to maintain.
- Elementary num theory Q: or
- Does $\lim_{n \rightarrow \infty} \frac{2^{n^2}}{n!}$ exist?
- Algebra Question 3
- What am I doing wrong?
- Derive and show
- Allocation of Price and Volume changes to a change in Rate
- Solving Inequalities- Erik and Nita are playing a game with numbers