Is this true that $\mathrm{Tr}(ABC)=\mathrm{Tr}(ACB)$? Let $A,B,C\in M_n$.
Is this true that $\mathrm{Tr}(ABC)=\mathrm{Tr}(ACB)$?
 A: When $n>1$, the equality is not true in general. The following counterexample works for all fields. Let
$$
A=\pmatrix{0&1\\ 0&0},\ B=A^T,\ C=\pmatrix{1&0\\ 0&0}.
$$
Then $ABC-ACB=C$, which is not traceless.
A: Try examples!
Let 
$$
\begin{align*}
A &= \begin{bmatrix}1&2\\ 3&4\end{bmatrix} &
B &= \begin{bmatrix}-1&2\\ 1&0\end{bmatrix} &
C &= \begin{bmatrix}1&0\\ 2&2\end{bmatrix} 
\end{align*}
$$
Then
$$
\begin{align*}
ABC &= \begin{bmatrix}5&4\\13&12\end{bmatrix} &
ACB &= \begin{bmatrix}-1 & 10\\-3&22\end{bmatrix}
\end{align*}
$$
Thus $\DeclareMathOperator{trace}{trace}$
$$
\begin{align*}
\trace(ABC) &= 17 & \trace(ACB) &= 21
\end{align*}
$$
This proves that the formula $$\trace(ABC)=\trace(ACB)$$ fails in general.
A: It is true that $\operatorname{tr}(ABC) = \operatorname{tr}(BCA) = \operatorname{tr}(CAB)$, but not that $\operatorname{tr}(ABC) = \operatorname{tr}(ACB)$.  Someone else has posted a counterexample to the latter proposition.
Forget square matrices; just suppose $A\in\mathbb R^{m\times n}$ and $B\in\mathbb R^{n\times m}$.  Then one can show
$$
\operatorname{tr}(AB) = \operatorname{tr}(BA). \qquad \text{(Note that $AB\in\mathbb R^{m\times m}$ and $BA\in\mathbb R^{n\times n}$.)}
$$
The result asserted above to be true follows from this.
Proof:
\begin{align}
& \operatorname{tr}(AB) = \sum_{i=1}^m (AB)_{ii} = \sum_{i=1}^m \sum_{j=1}^n A_{ij} B_{ji} \\[8pt]
= {} & \sum_{j=1}^n \sum_{i=1}^m B_{ji} A_{ij} = \sum_{j=1}^n (BA)_{jj} = \operatorname{tr}(BA). \qquad \blacksquare
\end{align}
