 Solution

If $\hat{A}|a⟩=\lambda |a⟩$, the Hermitian operator property makes $\lambda$ real. Applying the operator identity to this eigenvector gives ${\lambda }^3=1$. Its only real root is $\lambda =1$; the other cube roots of unity are nonreal. In an orthonormal eigenbasis, $\hat{A}={\sum }_a|a⟩⟨a|=\boxed{\hat{I}}$. Equivalently, the spectral theorem places the entire spectrum of the self-adjoint observable at one, giving the same conclusion. The quantum-mechanical setting assumes the usual well-defined observable and operator domains.