Not necessarily. It depends on A. If A is an "invertible" operator then we could apply the inverse to both sides, [tex]A^{-2}(A^3x)= (A^{-2}A^3)x= Ax= 0[/tex] to show that x must be 0. But if A is not invertible, then [tex]A^3x[/tex] can be 0 when Ax is not necessarily 0. For a counter example, consider [tex]A= \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}[/tex]. [tex]A^3[/tex] is the 0 matrix so [tex]A^3x= 0[/tex] for all x.