Question

Solve each inequality algebraically.

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

Short answer

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

Step-by-step solution

Step 1. Given information  

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

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

$$\begin{gathered} {(2-x)}^3=02-x=0 \\ x=2 \\ 3x-2=0 \\ 3x=2 \\ x=\frac{2}{3} \\ {x}^3+1=0 \\ {x}^3=-1 \\ x=-1 \end{gathered}$$

Step 3. Form the intervals  

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

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

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

Create the following table:

Interval	$(-\infty ,-1)$	$(-1,\frac{2}{3})$	$(\frac{2}{3},2)$	$(2,\infty )$
Number chosen	$-2$	0	1	3
Value of f	$f(-2)=\frac{512}{7}$	$f\left(0\right)=-16$	$f\left(1\right)=\frac{1}{2}$	$f\left(3\right)=-\frac{1}{4}$
Conclusion	positive	negative	positive	negative

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 localid="1646212202953" $(-1,\frac{2}{3})\cup (2,\infty )$.