x^3 - 4x < 0
x ( x^2 - 4) < 0
x (x - 2) (x + 2)< 0 (1)
We have 4 possible intervals that might work
(-inf, -2) or (-2, 0) or (0, 2) or ( 2, inf)
Note that if
x = -3 then the terms in (1) have the signs (-)(-)(-) = - so true for this interval
x = -1 then the terms in (1) have the signs ( -)(-))+) = + so not true for this interval
x = 1 then the signs are (+)(-)(+) = - so true for this interval
x =3 the signs are (+) (+)(+) = + so not true for this interva;
So.....the solution intevals are ( -inf, -2) U ( 0, 2)
See the graph here to confirm this : https://www.desmos.com/calculator/vyj5pxzrj5