Answer
$(-\infty,-1) \cup [1,\infty)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{x-1}{x^3+1} \geq 0$
$f(x)=\displaystyle \frac{(x-1)}{(x+1)(x^2-x+1)} \geq 0$
2. The cut points are:
$\displaystyle \displaystyle \frac{(x-1)}{(x+1)(x^2-x+1)} = 0$
$x=-1$ or $x=1$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\
& & \displaystyle \displaystyle \frac{(a-1)}{(a+1)(a^2-a+1)} & \\
(-\infty,-1) & -5 & \frac{(-)}{(-)(+)}=(+) & T\\
(-1, 1) & 0 & \frac{(-)}{(+)(+)}=(-) & F\\
(1,\infty) & 10 & \frac{(+)}{(+)(+)}=(+) & T\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\infty,-1) \cup [1,\infty)$
