Show that $\det(A^3+B^3)$ is the sum of two cubes of integers

Question

Let $ A,B$ be two integer matrices of dimension $3$ with $$AB=BA \ \text{ and } \ \det A= \det B$$ Show that $\det(A^3+B^3)$ is the sum of two cubes of integers.
Should I write the function so I have: $$\det(A^3-3AB+3BA-B^3)=\det(A-B)^3?$$ I don't know how to solve this. Any help?

Answer 1

At the time of writing, you have asked four questions (excluding this one) about linear algebra on this site. The problem statements of exactly half of them, which are supposed to be true, are actually false. In future, please refrain from throwing random conjectures at the users here, and try your best to make sure that the questions are correct before posting them.

Apparently, the problem statement of this question is false too. Let $A=\operatorname{diag}(1,1,2)$ and $B=\operatorname{diag}(2,1,1)$. Clearly, $AB=BA$ and $\det(A)=\det(B)$. However $n=\det(A^3+B^3)=162$ is not a sum of two cubic powers of integers. You may find a list of integers that can be written as sums of two cubic integer powers from OEIS sequence A045980. The number $162$ is not on the list. Alternatively, you may apply the results from the following paper:

Kevin A. Broughan (2003), Characterizing the Sum of Two Cubes, J. Int. Seq., v.6, article 03.4.6.

According to Broughan, given a positive integer $n$,

1. $n=x^3+y^3$ is solvable by positive integers $x,y$ if and only if
   • $n$ has a divisor $m$ with $n^{1/3} \le m \le (4n)^{1/3}$,
   • $l = \frac13\left(m^3 - \frac nm\right)$ is a positive integer, and
   • $s = m^2-4l$ is a perfect square.
2. $n=x^3+(-y)^3$ is solvable by positive integers $x,y$ if and only if
   • $n$ has a divisor $m$ with $1 \le m \le n^{1/3}$,
   • $l = \frac13\left(\frac nm - m^3\right)$ is a positive integer, and
   • $s = m^2+4l$ is a perfect square.

In our case, $n=162=2\times3^4$. The number itself is not a perfect cube. Therefore, if it can be written as the sum of two cubic powers of integers, both summands are nonzero and hence one of the two aforementioned cases in Broughan's paper holds.

For case 1, the first criterion implies that $6\le m\le8$ and hence $m=6$; hence $l=63$ and $s=-216$ is not a perfect square.

For case 2, the first criterion implies that $1\le m\le5$ and hence $m=2$ or $m=3$; the second criterion narrows down the choices to $(m,l)=(3,9)$; the third criterion shows that this choice is invalid, because $s=45$ is not a perfect square.

Therefore, $n=\det(A^3+B^3)=162$ is not a sum of two integer cubic powers.