Solution:
a) Let $x$ satisfy the equality $[x]^3 + x^2 = x^3 + [x]^2$. Then setting $t = [x]$ and $\alpha = x - t \in [0,1)$ one has that
$$t^3 - t^2 = (t + \alpha)^3 - (t + \alpha)^2 \Longleftrightarrow \alpha\left(\alpha^2 + (3t - 1)\alpha + 3t^2 - 2t\right) = 0$$
Hence either $\alpha = 0$, or $\alpha$ is a root of the equation in the brackets. In the second case the discriminant $(3t + 1)(1 - t)$ of this equation must be non-negative. Since $t$ is an integer, it follows that $t = 0$ or $t = 1$. Then either $\alpha = 0$, $\alpha = 1$ or $\alpha = -1$. Now $\alpha \in [0,1)$ implies that $\alpha = 0$, i.e., $x$ is an integer.

b) The degree of the polynomial $y^3 - y^2 - 1$ is odd. Hence this polynomial has a real zero $\alpha$. Obviously, $\alpha$ is not an integer (in fact, $\alpha$ is unique and $\alpha \in (1,2)$). Then $\left[\alpha^3\right] = \left[\alpha^2 + 1\right] = \left[\alpha^2\right] + 1$ and hence $\left[\alpha^3\right] - \left[\alpha^2\right] = 1 = \alpha^3 - \alpha^2$.