Cube and Square

$\text{Let the integers be} ~~~~(x-2)(x-1)x(x+1)(x+2). ~~~~~~A,~B ~\in \large ~Z^+,\\ so~ that~~ 5x=A^3...i) ~~~and~~~ 3x=B^2...ii)~~~~~~\implies \dfrac{A^3}{5}=x=\dfrac{B^2}{3}\\\text{This is possible only if} ~~\dfrac{3*A^3}{5}=B^2.\\\text{So if A contains 3, }3*3^3 = 9^{\color{#D61F06}{\large 2}} .\\ \text{If A contains 5, } \dfrac{5^3}{5}=5^{\color{#D61F06}{\large 2}}.\\\therefore~5X=3^3*5^3.~~\implies~x=27*25=675.\\\text{x is the Middle Number and 3, and 5 are the smallest possible integers.}\\\text{So the Middle Number has a value of }~~~~~~~~\color{#3D99F6}{\large \boxed{675} }\\~~~~\\\text{I got this value first through TI-83 calculator by repeatedly getting}\\ 1+x ~~STR~~ x:~~\sqrt{\dfrac{3}{5} *x^3}\\ \text{ till I got an integer as the result. x is set to 0 to start with.}$