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The problem is as follows:

There are two cubes, a big one of which we know the edge is $$2x^2+1$$ and a small one with a volume of $$x^2-1.$$

If we try to fill the big cube with cubes like the small one we get a remaining space that has exactly the same volume of the small cube.

How long is the edge of the bigger cube?

Edit 1: I invented the problem to be solved using Ruffini's rule, but I would like to have some feedback about it :)

Edit 2: removed the dimensions by sugestion of @okrzysik

NOTE: Instead of having the remaning space equal to the volume of the small cube, there should be a difference, so assume that the remaning volume has less 4 cubic units than the smaller cube.

Concept7
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  • What dimensions does $x$ have? – okrzysik Nov 20 '15 at 00:42
  • @okrzysik it is not given, you have to find it. Can you do that? – Concept7 Nov 20 '15 at 00:45
  • I don't think you understand what I mean, does $x$ have the dimension of length or length$^{1/2}$ or something? Or is $x$ dimensionless? – okrzysik Nov 20 '15 at 00:49
  • oh, yes, I didn't understand. It's dimensionless, just a number. – Concept7 Nov 20 '15 at 00:51
  • OK cool. It doesn't really make sense for you to have the cm and the cm$^3$ in the question then. If $x$ is dimensionless, then $x^2$ also has to be dimensionless, so it cannot have the dimension of length, let alone have units of cm. You need to state the cube has a dimensionless length of "...", and a dimensionless volume of "...". – okrzysik Nov 20 '15 at 00:55
  • @okrzysik, I agree , units are irrelevant. :) – Concept7 Nov 20 '15 at 01:00

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Well, if the remaining space in the packed big cube is exactly the volume of one small cube, then we could technically fit an extra small cube inside the big cube. So, we really just have to find how many small cubes can we fit inside our big cube.

Since the big cube's edge is $2x^2+1$, then the volume of the big cube is $(2x^2+1)^3$. Then, the number of small cubes we can fit inside our big cube is just $$\frac{(2x^2+1)^3}{x^2-1} = \frac{8 x^6+12 x^4+6 x^2+1}{x^2-1} = 8x^4 +20x^2 +26 + \frac{27}{x^2-1}.$$

Setting the above to an integer $k$, we now have to solve for the zeros of $$(8 x^4+20 x^2+26-k)(x^2-1)+27.$$ That is, the zeros of $$8 x^6+12 x^4+6 x^2 -k x^2 + k +1 = 0.$$

But $k$ is variable, so there's many solutions to this.

With the aid of a CAS though, we find integer roots $x=\pm 2$ at $k=243$. Are there any others?

Here the edge of the big cube is $9cm$, and we can fit exactly $243$ cubes of volume $3cm^3$ into it.

This is a cool problem.

Andrey Kaipov
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  • first of all, thanks for the feedback. Why did you solve $$8x^6+12x^4+6x^2-kx^2+k+1=0$$ and not $$x^2-1=27$$? It is said that the remaining space has the same volume as the small cube. – Concept7 Nov 20 '15 at 00:38
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    I'm sorry I don't understand. Did I over complicate something? Where did $x^2-1=27$ come from? I now see that if we want $x$ to be an integer, then we could also solve $27 = k(x^2-1)$. That makes it easier! But we're leaving out the noninteger solutions like $x=\sqrt{7}/2$ at $k=243/2$ if we do that. – Andrey Kaipov Nov 20 '15 at 00:57
  • ok, don't tell me I'm failing at my own problem... :)

    I'm thinking like this:

    $$8x^6+12x^4+6x^2+1= (8x^4+20x^2+26)(x^2-1)+27$$

    So, there are $8x^4+20x^2+26$ cubes like the smaller one and the remaining volume is 27 so $x^2-1$ which is the volume of the small cube is equal to 27. Isn't it?

     – Concept7 Nov 20 '15 at 01:20
  • Oh man, your way makes sense to me too. I've been thinking for a while for some correspondence between the two, but it's making me flustered. It's a fun problem! – Andrey Kaipov Nov 20 '15 at 03:17
  • I had to change the problem, see note in original post, since the cube would be perfectly divided in smaller cubes and no space would remain. :/ – Concept7 Nov 20 '15 at 12:37