Prove that $\operatorname{trace}(ABC) = \operatorname{trace}(BCA) = \operatorname{trace}(CAB)$ [closed]

Is it already known that $\operatorname{Tr}(XY) = \operatorname{Tr}(YX)$ when $X$ and $Y$ are square matrices of the same size?

If it is, then simply set $X= AB$ and $Y = C$. It will give you $\operatorname{Tr}(ABC) = \operatorname{Tr}(CAB)$. You can get $\operatorname{Tr}(ABC) = \operatorname{Tr}(BCA)$ in a similar fashion.

Hint

$$tr(ABC)=\sum_i (ABC)_{ii}=(ABC=A(BC))=\sum_i\sum_j A_{ij}(BC)_{ji}= \sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki};$$

now you can exchange the order of the matrices to arrive at the thesis as each of the $A_{ij}$, $B_{jk}$ and $C_{ki}$ are scalars (considering matrices over $\mathbb R$, for example). We arrive at

$$tr(ABC)=\sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki}=\sum_i\sum_j\sum_kB_{jk}C_{ki}A_{ij}=(BCA=(BC)A)=tr(BCA), $$

and so on.

Hint: use the definition of trace.

$$\text{Tr}(ABC)=\sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki}.$$