= Solution
For $F(x)=x(\mu-x^2)$, the <fixed points> are zero for all $\mu$ and $\pm\sqrt{\mu-1}$ for $\mu>1$. Their multipliers are respectively $\mu$ and $3-2\mu$, giving attraction for $-1<\mu<1$ and $1<\mu<2$ respectively.
Remove these <fixed points> from the roots of $F^2(x)=x$. The factor $\mu+1-x^2$ gives the genuine two-cycle
$$
\boxed{\{+\sqrt{\mu+1},-\sqrt{\mu+1}\},\qquad\mu>-1.}
$$
Its multiplier is $(2\mu+3)^2>1$, so it is always unstable. It collapses to zero at $\mu=-1$, the subcritical <period-doubling bifurcation> of zero.
The remaining equation is $x^4-\mu x^2+1=0$, whose positive roots in $x^2$ exist only for $\mu\geq2$. At $\mu=2$ they give the <fixed points> $\pm1$, not genuine two-cycles. For $\mu>2$, put $r=\sqrt{(\mu+\sqrt{\mu^2-4})/2}>1$. Since $\mu=r^2+r^{-2}$, $F(r)=r^{-1}$ and $F(r^{-1})=r$. There are exactly two further cycles,
$$
\boxed{\{r,r^{-1}\},\qquad\{-r,-r^{-1}\},\qquad\mu>2.}
$$
Their common multiplier is $(\mu-3r^2)(\mu-3r^{-2})=9-2\mu^2$. They are attracting for $2<\mu<\sqrt5$ and repelling for $\mu>\sqrt5$, with marginal multiplier $-1$ at $\sqrt5$. Thus there are no cycles for $\mu\leq-1$, one for $-1<\mu\leq2$, and three for $\mu>2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4-two-cycles.png]
{title=Fixed points and two-cycle points for the odd cubic map}
The nonzero <fixed points> emerge in a supercritical <pitchfork bifurcation> at one and each flips to an attracting two-cycle at two. Beyond $\sqrt5$ further period doubling to four-cycles is expected, followed by additional bifurcations and potentially chaotic dynamics. This local expectation does not assert chaos for every larger parameter; periodic windows and escape are possible.
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