= Solution
Choose a basis whose first axis is $\mathbf n$. The tensor with <uniaxial nematic order> then has <eigenvalues> $2Q,-Q,-Q$. Hence $\operatorname{Tr}S^2=6Q^2$, $\operatorname{Tr}S^3=6Q^3$ and $\operatorname{Tr}S^4=18Q^4$. Substitution gives
$$
\boxed{f(Q)=6aQ^2+6bQ^3+18cQ^4.}
$$
For $a,c>0$, $Q=0$ remains a <local minimum> with curvature $12a$. Its nonzero <stationary points> satisfy $2a+3bQ+12cQ^2=0$, so
$$
Q_\pm=\frac{-3b\pm\sqrt{9b^2-96ac}}{24c}.
$$
They first appear at $b_s=-\sqrt{32ac/3}$ as a stationary inflection point; for more negative $b$, $Q_-$ is a <local maximum> and $Q_+$ a <local minimum>. The latter is initially metastable, becomes degenerate with $Q=0$ at $b_t=-\sqrt{12ac}$, and then becomes the <global minimum> as $b$ decreases further. This is a <first-order transition in a cubic-quartic Landau potential>: \b[the order changes discontinuously]. The <saddle-node bifurcation> creating a <local minimum> is not the coexistence condition. The isotropic minimum never loses local stability in this fixed positive-$a$ scan.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2016/paper-1-nematic-free-energy.png]
{title=Cubic nematic invariant creates a metastable minimum followed by first-order coexistence}
The illustrative plots set $a=c=1$ and show $f/6$. A cubic invariant is permitted because head-tail reversal leaves $S$ unchanged; it does not implement $S\mapsto-S$. The <physical range of a uniaxial nematic order parameter> imposes $-1/2\leq Q\leq1$; the usual small-order-parameter Landau result assumes the displayed minimum lies in that interval. The unrestricted quartic graphs express the intended Landau model, with the effect of enforcing the bound stated in part (f).<Lectures on Landau Theory of Phase Transitions by Peter D. Olmsted>[§4.3, printed pp.11–13, tensor invariants and cubic-term-driven coexistence; coefficient normalization here is derived from this paper's definition of $S$]{ref}
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