= Solution
If $\hat A|a\rangle=\lambda|a\rangle$, the <Hermitian operator> property makes $\lambda$ real. Applying the operator identity to this <eigenvector> gives $\lambda^3=1$. Its only real root is \b[$\lambda=1$]; the other cube roots of unity are nonreal. In an orthonormal <eigenbasis>, $\hat A=\sum_a|a\rangle\langle a|=\boxed{\hat I}$. 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.
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