If the function $f$ given by $f(x)=x^3-3(a-2)x^2+3ax+7$, for some $a\in\mathbb{R}$, is increasing in $(0,1]$ and decreasing in $[1,5)$, then a root of the equation $\dfrac{f(x)-14}{(x-1)^2}=0\ (x\ne1)$ is: