The cubic can have up to three zeroes. In fact, it has only one real zero and two complex zeroes. f(0)=1 and f(1)=1, but f(-1)=-1, so there must be a real zero between x=0 and x=-1.
We can use Newton's method to find the real zero.
xn+1=xn-f(xn)/f'(xn).
f'(x)=3x2-2x. We need to choose a suitable x0. We can't use x0=0 because it would lead to division by zero, so let x0=-1.
x1=-1-f(-1)/f'(-1)=-4/5, x2=-333/440, x3=-0.7548814744, x4=-0.7548776663, x5=-0.7548776662, (x6=-0.754877666247). The fifth iteration after x0=-1 is stable so is an approximate solution for the real zero of f(x).