Given the definitions
$$ P_\mu= -i\partial_\mu $$ $$ Q_\alpha=-i(\partial_\alpha-(\sigma^\mu\bar{\theta})_\alpha\partial_\mu) $$ $$ \bar{Q_\dot{\alpha}}=+i(\bar{\partial}_\dot{\alpha}-({\theta}\sigma^\mu)_\dot{\alpha}\partial_\mu) $$ And the supersymmetric algebra which these satisfy $$ \{Q_\alpha,\bar{Q}_\dot{\alpha}\}=2\sigma^\mu_{\alpha\dot{\alpha}}P_\mu\;\;\;\;\{Q_\alpha,{Q}_{\beta}\}=\{\bar{Q}_\dot{\alpha},\bar{Q}_\dot{\beta}\}=0 $$
How can I show that $$ [\epsilon_1Q+\bar{\epsilon_1}\bar{Q},\epsilon_2Q+\bar{\epsilon_2}\bar{Q}] = 2(\epsilon_1\sigma^\mu\bar{\epsilon_2}-\epsilon_2\sigma^\mu\bar{\epsilon_1})P_\mu $$
given that $\epsilon_1,\epsilon_2$ are Grassmann odd spinor supersymmetry parameters?
I have started by decomposing the commutator to $$ [\epsilon_1Q+\bar{\epsilon_1}\bar{Q},\epsilon_2Q+\bar{\epsilon_2}\bar{Q}] = [\epsilon_1Q,\epsilon_2Q]+[\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q]+[\bar{\epsilon_1}\bar{Q},\bar{\epsilon_2}\bar{Q}] $$ We know from the supersymmetric algebra that the first and last commutators above are zero which gives $$ [\epsilon_1Q+\bar{\epsilon_1}\bar{Q},\epsilon_2Q+\bar{\epsilon_2}\bar{Q}] = [\epsilon_1Q,\bar{\epsilon_2}\bar{Q}]+[\bar{\epsilon_1}\bar{Q},\epsilon_2 Q] $$ From what we are given I suspect that I have to turn the commutators into anti-commutators but I'm not sure how I can do that here.