Answer
Solution set: $(-\infty, -1)( \cup \{1\} $
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.
$\displaystyle \frac{x^2-2x+1}{x^3+3x^2+3x+1} \leq 0$,
$\displaystyle \frac{(x-1)(x-1)}{(x+1)(x+1)(x+1)} \leq 0$
$f(x)=\displaystyle \frac{(x-1)(x-1)}{(x+1)(x+1)(x+1)} \leq 0$,
2.The cut points are:
$\displaystyle \frac{(x-1)(x-1)}{(x+1)(x+1)(x+1)} =0$
$x=-1$ 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,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& & \displaystyle \displaystyle \frac{(a-1)(a-1)}{(a+1)(a+1)(a+1)} & \\
(-\infty,-1) & -5 & \frac{(-)(-)}{(-)(-)(-)}=(-) & T\\
(-1, 1) & 0 & \frac{(-)(-)}{(+)(+)(+)}=(+) & F\\
(1,\infty) & 5 & \frac{(+)(+)}{(+)(+)(+)}=(+) & F\\
\end{array}$
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 unless they belong to the denominator.
Solution set: $(-\infty, -1) \cup \{1\} $
