jmc

We are asked to find the integers whose cubes are between $-50$ and $50$. Since $f(x)=x^3$ is a monotonically increasing function, we can find the least and the greatest integers satisfying the inequality and count the integers between them, inclusive (see graph). Since $3^3=27<50$ and $4^3=64>50$, $n=3$ is the largest solution. Similarly, $n=-3$ is the smallest solution. Therefore, there are $3-(-3)+1=\boxed{7}$ solutions.