Let $m_1\otimes m_2\otimes m_3\in M\otimes M\otimes M$, then $ \begin{align} R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1\otimes m_2\otimes m_3)&= (\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(1_M\otimes\tau)(m_1\otimes m_2\otimes m_3)\\ &=(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(m_1\otimes m_3\otimes m_2)\\ &=(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(m_1\otimes m_2\otimes m_3)\\ &=(\tau\otimes 1_M)(1_M\otimes\tau)(m_2\otimes m_1\otimes m_3)\\ &=(\tau\otimes 1_M)(m_2\otimes m_3\otimes m_1)\\ &=m_3\otimes m_2\otimes m_1\\ \end{align} $ $ \begin{align} R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1\otimes m_2\otimes m_3)&= (1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(\tau\otimes 1_M)(m_1\otimes m_2\otimes m_3)\\ &=(1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(1_M\otimes\tau)(m_2\otimes m_1\otimes m_3)\\ &=(1_M \otimes \tau)(1_M\otimes\tau)(\tau\otimes 1_M)(m_2\otimes m_3\otimes m_1)\\ &=(1_M \otimes \tau)(1_M\otimes\tau)(m_3\otimes m_2\otimes m_1)\\ &=(1_M \otimes \tau)(m_3\otimes m_1\otimes m_2)\\ &=m_3\otimes m_2\otimes m_1\\ \end{align} $ so we conclude $ R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1\otimes m_2\otimes m_3)=R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1\otimes m_2\otimes m_3)\tag{1} $ Now take arbitrary $u\in M\otimes M\otimes M$, then we have representation $ u=\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)} $ Hence using $(1)$ we get $ \begin{align} R_{(1,2)}R_{(1,3)}R_{(2,3)}(u) &=R_{(1,2)}R_{(1,3)}R_{(2,3)}\left(\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)}\right)\\ &=\sum\limits_{i=1}^n R_{(1,2)}R_{(1,3)}R_{(2,3)}(m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)})\\ &=\sum\limits_{i=1}^n R_{(2,3)}R_{(1,3)}R_{(1,2)}(m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)})\\ &=R_{(2,3)}R_{(1,3)}R_{(1,2)}\left(\sum\limits_{i=1}^n m_1^{(i)}\otimes m_2^{(i)}\otimes m_3^{(i)}\right)\\ &=R_{(2,3)}R_{(1,3)}R_{(1,2)}(u)\\ \end{align} $ Since $u\in M\otimes M\otimes M$ is arbitrary we conclude $ R_{(1,2)}R_{(1,3)}R_{(2,3)}=R_{(2,3)}R_{(1,3)}R_{(1,2)} $