$\text{ Let }a>2\text{ be a constant. If there are just }18\text{ positive integers satisfying the inequality }(x-a)(x-2a)\left(x-a^2\right)<0\text{ then the value of }a\text{ is }$

$\begin{array}{r}\text{ as }a>2\text{ hence }\\ a^2>2a>a>2\end{array}$

$\text{ now }(x-a)(x-2a)\left(x-a^2\right)<0\Rightarrow \text{ the solution set is as shown }$

$\begin{array}{l}\text{ between }(0,a)\text{ there are }(a-1){\text{ positive integers and between (2a,a}}^2{\text{), there are a}}^2\text{ -1-2a positive integers}\\ \therefore a^2-1-2a+a-1=18\\ \Rightarrow a^2-a-20=0\\ \Rightarrow (a-5)(a+4)=0\\ \therefore a=5\end{array}$