Let $R$ be an associative ring with identity $1$ and $J\left(R\right)$ the Jacobson radical of $R$. Suppose that $m\ge 1$ is a fixed positive integer and $R$ an $m$-torsion-free ring with $1$. In the present paper, it is shown that $R$ is commutative if $R$ satisfies both the conditions (i) $[x^m,y^m]=0$ for all $x,y\in R\setminus J\left(R\right)$ and (ii) $[x,[x,y^m]]=0$, for all $x,y\in R\setminus J\left(R\right)$. This result is also valid if (ii) is replaced by (ii)’ $[{\left(yx\right)}^m{x}^m-x^m{\left(xy\right)}^m,x]=0$, for all $x,y\in R\setminus N\left(R\right)$. Our results generalize many well-known commutativity theorems (cf. [1], [2], [3], [4], [5], [6], [9], [10], [11] and [14]). ...

In this paper we investigate commutativity of rings with unity satisfying any one of the properties: $$\begin{array}{cc}& \{1-g\left(y{x}^m\right)\}[y{x}^m-x^{r}f\left(y{x}^m\right)x^s,x]\{1-h\left(y{x}^m\right)\}=0,\hfill \\ & \{1-g\left(y{x}^m\right)\}[x^{m}y-x^{r}f\left(y{x}^m\right)x^s,x]\{1-h\left(y{x}^m\right)\}=0,\hfill \\ & y^t[x,y^n]=g\left(x\right)[f\left(x\right),y]h\left(x\right)\mathrm{a}nd[x,y^n]y^t=g\left(x\right)[f\left(x\right),y]h\left(x\right)\hfill \end{array}$$ for some $f\left(X\right)$ in $X^2\mathbb{Z}\left[X\right]$ and $g\left(X\right)$, $h\left(X\right)$ in $\mathbb{Z}\left[X\right]$, where $m\ge 0$, $r\ge 0$, $s\ge 0$, $n>0$, $t>0$ are non-negative integers. We also extend these results to the case when integral exponents in the underlying conditions are no longer fixed, rather they depend on the pair of ring elements $x$ and $y$ for their values. Further, under different appropriate constraints on commutators, commutativity of rings has been studied. These results...

Let $m>1,s\ge 1$ be fixed positive integers, and let $R$ be a ring with unity $1$ in which for every $x$ in $R$ there exist integers $p=p\left(x\right)\ge 0,q=q\left(x\right)\ge 0,n=n\left(x\right)\ge 0,r=r\left(x\right)\ge 0$ such that either $x^p[x^n,y]x^q=x^r[x,y^m]y^s$ or $x^p[x^n,y]x^q=y^s[x,y^m]x^r$ for all $y\in R$. In the present paper it is shown that $R$ is commutative if it satisfies the property $Q\left(m\right)$ (i.e. for all $x,y\in R,m[x,y]=0$ implies $[x,y]=0$).