本文解决了以下三加性s-泛函不等式egin{eqnarray}label{0.1}&onumberf(x+y,z-w,a+b)+f(x-y,z+w,a-b)\&onumberqquad-2f(x,z,a)+2f(x,w,b)-2f(y,z,b)+2f(y,w,a)\&quadleftsleft(2fleftfleft(frac{x+y}{2},z-w,a+b ight)+2fleft(frac{x+y}{2},z-w,a+b ight)+2fleft(frac{x-y}{2},z+w,a+b ight).轻点。\&qquad左。左。-2 f(x,z,a)+2 f(x,w,b)-2 f(x,w,b)+2 f(y,w,a)轻,onumber结束{eqnarray}开始{eqnarray}标签{0.1}&onumberf(x+y,z-w,a+b)+f(x,z,a)+2 f(x,w,b)-2f(y,z,b)+2 f(y,w,a)\\&四lefsleft(2fleft(frac{x+y}{2},z)-w,a+b光)+2f左(FRAC{x-y}{2},z+w,a-光)光。轻点。\&qquad左。左。-2 f(x,z,a)+2 f(x,w,b)-2f(y,z,b)+2 f(y,w,a)轻点,onumber结束{eqnarray}开始{eqnarray}标签{0.2}&onumber左2fleft(frac{x+y}{2},z-w,a+b轻点)+2fleft(frac{x+y}{2},z+w,a-bight)轻点。\&&onumberqquadleft。-2 f(x,z,a)+2 f(x,w,b)-2f(y,z,b)+2f(y,w,a)轻\&四les(f(x+y,z-w,a+b)+f(x-y,z+w,a-b)\&onumberqquad-2 f(x,z,a)+2 f(x,w,b)-2f(y,z,b)+2f(y,w,a))\结束{eqnarray}开始{eqnarray}标签{0.2}&onumberleftfrac{x+y}{2},z-w,a+b右)+2f左(frac{x-y}{2},z+w,a-bight)右。\&&onumberqquadleft。-2 f(x,z,a)+2 f(x,w,b)-2f(y,z,b)+2 f(y,w,a)轻\&四les(f(x+y,z-w,a+b)+f(x-y,z+w,a-b)\&一个数字q四-2 f(x,z,a)+2 f(x,w,b)-2f(y,z,b)+2 f(y,w,b))\end{eqnarray}其中s<1s<1的固定非零复数。此外,我们证明了Banach代数和单元C^*C^*-代数中置换三导子和置换三同态的Hyers-Ulam稳定性和超稳定性,并与三可加s s-泛函不等式{m(ef{0.1}ef{0.1})}和{m(ef{0.2}ef{0.2})}联系起来。