Five. (20 points) Suppose a line intersects the graph of the function $y=x^{4}-x^{2}+x$ at exactly two distinct points. Find all such line equations.

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Five. (20 points) Suppose a line intersects the graph of the function $y=x^{4}-x^{2}+x$ at exactly two distinct points. Find all such line equations.
Obviously, the line $x=a$ intersects the graph of the function $y=x^{4}-x^{2}+x$ at only one point. Therefore, let the equation of the line be $y=p x+q$. Substituting it into $y=x^{4}-x^{2}+x$, we get
$$
x^{4}-x^{2}+(1-p) x-q=0 \text {. }
$$

Equation (1) has exactly two distinct real roots, which can occur in the following 3 cases:
$$
\begin{array}{l}
\text { (1) } x^{4}-x^{2}+(1-p) x-q \\
=(x-u)(x-v)\left(x^{2}+C x+D\right),
\end{array}
$$

where $u, v, C, D \in \mathbf{R}, u \neq v$, and $C^{2}4 B, C^{2}4 B, C^{2}\sqrt{2} .\end{array}\right.$
Then $p=1-B C-A D=A+2 A B+1-A^{3}$,
$$
q=-B D=B+B^{2}-A^{2} B \text {. }
$$

The equation of the line is
$$
y=\left(A+2 A B+1-A^{3}\right) x+B+B^{2}-A^{2} B,
$$

where the real numbers $A, B$ satisfy $B4 B$.
In the system of equations (1), let $A=C, B=D$, we get
$$
2 A=0, A^{2}+2 B=-1,2 A B=1-p, B^{2}=-q \text {. }
$$

Solving these, we get $A=0, B=-\frac{1}{2}, p=1, q=-\frac{1}{4}$.
Therefore, the equation of the line is $y=x-\frac{1}{4}$.
For (3), expanding and comparing coefficients, we get
$$
\begin{array}{l}
3 u+v=0,3\left(u^{2}+u v\right)=-1, \\
u^{3}+3 u^{2} v=p-1, u^{3} v=-q .
\end{array}
$$

From the first two equations, we get
$$
v=-3 u, 3\left(u^{2}-3 u^{2}\right)=-1 \text {. }
$$

Solving these, we get $u= \pm \frac{\sqrt{6}}{6}$.
Noting that,
$$
\begin{array}{l}
p=1+u^{2}(u+3 v)=1+\frac{1}{6}(u-9 u)=1-\frac{4}{3} u, \\
q=-u^{3} v=3 u^{4}=\frac{1}{12},
\end{array}
$$

we have $(p, q)=\left(1 \pm \frac{2 \sqrt{6}}{9}, \frac{1}{12}\right)$.
Thus, the equation of the line is
$$
y=\left(1 \pm \frac{2 \sqrt{6}}{9}\right) x+\frac{1}{12} .
$$