How to show the following problem of determinant

Question

If the matrices $A,B\in M_{3}(\Bbb{Z})$ are singular and $AB=BA$, show that the number $$\det(A^3+B^3)+\det(A^3-B^3)$$ is the double of a perfect cube.

I have considered the polynomial $$\det(A+xB)=\det A+mx+nx^2+x^3\det B$$ From this how I can show

Answer 1

Following Anurag A's suggestion in the comments, let $\omega$ be a primitive third root of unity (so $\omega^2+\omega + 1=0$).

Then observe that since $A$ and $B$ commute, we have $$A^3+B^3 = (A+B)(A+\omega B)(A+\omega^2B)$$ and $$A^3-B^3 = (A-B)(A-\omega B)(A-\omega^2 B).$$

Then using that $\det(A)=\det(B)=0$, we have that $p(x)=\det(A+xB)= x(m+nx)$ for some integers $n$ and $m$ from the formula that you've written in the question.

Thus $$\det(A^3+B^3)+\det(A^3-B^3) = p(1)p(\omega)p(\omega^2) +p(-1)p(-\omega)p(-\omega^2) $$ $$= (m+n)(m+n\omega)(m+n\omega^2) - (m-n)(m-n\omega)(m-n\omega^2) = m^3+n^3 - (m^3-n^3) = 2n^3.$$