Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 148: 93

$\{0,31\}$

Step 1. Let $u=(x+1)^{1/5}$, we have $u^2-3u+2=0$ Step 2. Factor and solve: $(u-2)(u-1)=0$ thus $u=1,2$ Step 3. For $u=1$, we have $(x+1)^{1/5}=1$ or $x+1=1$ thus $x=0$ Step 4. For $u=2$, we have $(x+1)^{1/5}=2$ or $x+1=32$ thus $x=31$ Step 5. Solution set $\{0,31\}$