Need a hint on a matrix problem We have two, $n \times n$ matrices $X, Y$  with complex elements which can be written like $X=A^3+B^3$ and $Y=C^3+D^3$, where $A,B,C,D$ are, respectively, matrices of the same dimension with complex elements. I have to show that there exist two matrices $E,F$ such that $XY=E^3+F^3$. Can someone, please, give me a hint?
 A: This question was asked in a recent contest. I hope the competition is now closed.
EDIT.  Proposition. Let $\alpha\in(1,+\infty)$ and let $A\in M_n(\mathbb{C})$. Then there are $E,F\in M_n(\mathbb{C})$ s.t. $E$ is a polynomial in $A$, $F$ is a scalar matrix and s.t. $A=E^{\alpha}+F^{\alpha}$.
Proof. Case 1. $A$ has no $\leq 0$ eigenvalues. Let $E=e^{(1/\alpha)\log(A)}$, where $\log$ is the principal logarithm. Then $E$ is a polynomial in $A$ and $E^{\alpha}=A$, because when $\alpha>1$, $(U^{1/\alpha})^{\alpha}=U$.
Case 2. $A$ admits $\lambda_1\leq\cdots\leq\lambda_k$ as $\leq 0$ eigenvalues; then $B=A+(1-\lambda_1)I_n$ has no $\leq 0$ eigenvalues and we may write $B=E^{\alpha}$, where $E$ is a polynomial in $B$, then in $A$. Finally $A=E^{\alpha}-(1-\lambda_1)I=E^{\alpha}+(\beta I)^{\alpha}$ where $1-\lambda_1=\rho$ and $\beta=\rho^{1/\alpha} e^{i\pi/\alpha}$.
A: Any complex matrix can be written as $A + iC$ where $A$ and $C$ are Hermitian. Hermitian matrices have cube roots (since they are diagonalizable) and so does $i$.
