Let $A,B,C$ be unitary matrices. Does there always exist a unitary matrix $X$ such that $$(XA)(XB)(XC)=I,$$ where $I$ is the identity matrix? The quadratic equation $(XA)(XB)=I$ has the solution $A^\*(AB^\*)^{1/2}$, and I am hoping that the cubic and higher dimensional versions are always solvable.



Matrix equation $XAXBXC=I$
Here is an argument showing that the answer is 'yes'. I'll let you check the details and that this result generalizes to all higher degrees. 


Consider the map $f\_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f\_{ABC}(X) = XAXBXC.
$$
Since the image of this map is compact, if this map were not onto, it would have to have topological degree equal to zero.


Next, since $\mathrm{U}(n)$ is connected, the map $f\_{ABC}:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by
$$
f\_{ABC}(X) = XAXBXC
$$
is homotopic to the map $f\_{III}$, and hence has the same degree as $f\_{III}$. Thus, it suffices to show that the map $f\_{III}$ has nonzero topological degree to show that $f\_{ABC}$ is surjective.


I claim that the map $f\_{III}$ has topological degree $3^n$. 


To prove this, it suffices to compute the local degrees around the pre-images of a regular value. Let $Y = \mathrm{diag}(e^{i\theta\_1},\ldots,e^{i\theta\_n})$, where $0<\theta\_1<\theta\_2<\cdots<\theta\_n<\pi$. Then $Y$ has distinct eigenvalues. Hence, any solution $X$ to $X^3 = Y$ has distinct eigenvalues and has the same eigendirections as $Y$. Thus, $X$, too, must be diagonal and must be of the form
$$
X = \mathrm{diag}(e^{i\tau\_1},\ldots,e^{i\tau\_n})
$$
where $3\tau\_k \equiv \theta\_k\ \mathrm{mod}\ 2\pi$ for $k=1,\ldots, n$. Thus, there are $3^n$ solutions $X$ to $X^3=Y$. 


I claim that $Y$ is a regular value of the map $f\_3:\mathrm{U}(n)\to \mathrm{U}(n)$ defined by $f\_3(X)=X^3$ and that $f\_3$ is orientation preserving at each solution $X$ of $X^3=Y$. This follows from the following computation: Consider the pullback under $f\_3$ of the canonical left-invariant form $g^{-1}\mathrm{d} g$.
$$
f\_3^\*(g^{-1}\mathrm{d}g) = g^{-3}\mathrm{d}(g^3) 
= (I + \mathrm{Ad}(g^{-1})+\mathrm{Ad}(g^{-2}))(g^{-1}\mathrm{d}g)
$$
When one computes the determinant of $\bigl(I + \mathrm{Ad}(X^{-1})+\mathrm{Ad}(X^{-2})\bigr):{\frak{u}}(n)\to {\frak{u}}(n)$ for each solution $X$ of $X^3=Y$, one finds that, because $Y$ has distinct eigenvalues, this determinant is a positive number. Thus, $Y$ is a regular value of $f\_3$, and each of the $3^n$ solutions $X$ to $X^3=Y$ contributes a $+1$ to the topological degree in the usual degree formula.


(By a similar argument, the map $f\_k:\mathrm{U}(n)\to\mathrm{U}(n)$ defined by $f(X) = X^k$ has topological degree $k^n$, which is nonzero, so it is necessarily surjective. This answers the higher degree cases as well.)


