Three consecutive numbers represented by n-1, n, n+1.
The product is n(n^2-1)=210 or n^3=210+n.
We can write this: n=210/(n^2-1).
Now we work out the nearest perfect cube to 210. 6^3=216 so pick 6.
n=210/(36-1)=210/35=6. But we started with 6 and got 6 again so n=6 and n-1=5, n+1=7.
The consecutive numbers are 5, 6, 7.