$(a+b) \cdot (a-b)=a^2-b^2, \forall a,b \in R \text{ iff R is commutative }$ Let $R$ a ring. Show that $(a+b) \cdot (a-b)=a^2-b^2, \forall a,b \in R \text{ iff R is commutative }$.
That's what I have tried:
$$(a+b) \cdot (a-b)=a^2-b^2 \Rightarrow a \cdot a+a(-b)+ba+b(-b)=a^2-b^2 \Rightarrow a^2-ab+ba-b^2=  a^2-b^2 \Rightarrow -ab+ba=0_R \Rightarrow ba=ab, \text{ so R is commutative}$$
Could you tell me if it is right??

If $\;R\;$ is commutative then
$$(a-b)(a+b)=a^2+\overbrace{ab-ba}^{=0\;, \text{ by comm.!}}-b^2=a^2-b^2$$
