Question

Solve each inequality algebraically.

$\frac{{(3-x)}^3(2x+1)}{x^3-1}<0$

Short answer

Solution to the inequality $\frac{{(3-x)}^3(2x+1)}{x^3-1}<0$is $(-\frac{1}{2},1)\cup (3,\infty )$.

Step-by-step solution

Step 1. Given information  

We have been given an inequality $\frac{{(3-x)}^3(2x+1)}{x^3-1}<0$.

We have to solve this inequality algebraically.

Step 2. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined.  

Assume $f\left(x\right)=\frac{{(3-x)}^3(2x+1)}{x^3-1}$.

$$\begin{gathered} {(3-x)}^3=0x=3 \\ 2x+1=0 \\ x=-\frac{1}{2} \\ {x}^3-1=0 \\ x=1 \end{gathered}$$

Step 3. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

$(-\infty ,-\frac{1}{2})\cup (-\frac{1}{2},1)\cup (1,3)\cup (3,\infty )$

Step 4. Select a number in each interval and evaluate f at the number  

Create the following table:

Interval	$(-\infty ,-\frac{1}{2})$	$(-\frac{1}{2},1)$	$(1,3)$	$(3,\infty )$
Number chosen	$-1$	0	2	4
value of f	$f(-1)=32$	$f\left(0\right)=-27$	$f\left(2\right)=\frac{5}{7}$	$f\left(4\right)=-\frac{1}{7}$
Conclusion	positive	negative	positive	nrgative

Step 5. Identify the interval 

Since we want to know where f is negative, we conclude that $f\left(x\right)<0$in the interval $(-\frac{1}{2},1)\cup (3,\infty )$.