= Solution
For a <normalizable> state take $a>0$. <inverse construction of a quantum potential> gives $V=E+\hbar^2\Psi''/(2m\Psi)$. <differentiation> yields $\Psi''/\Psi=16a^2x^6-12ax^2$, so the chosen <potential energy> zero fixes $E=0$. Thus
$$
\boxed{V(x)=\frac{8\hbar^2a^2}{m}x^6-\frac{6\hbar^2a}{m}x^2,\qquad E=0.}
$$
This <energy> lies above the negative minima of the <potential energy>; its zero value does not mean zero <kinetic energy>. The <nodeless eigenfunction> is the <ground state>.
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