Bài giải:

0,125(x + 1)³ – 1 = 0

 $\frac{1}{8}{{(x+1)}^{3}}-1$ = 0

 ${{\left[ \frac{1}{2}(x+1) \right]}^{3}}-1$ = 0

$\left[ \frac{1}{2}(x+1)-1 \right]\left[ \frac{1}{4}{{(x+1)}^{2}}+\frac{1}{2}(x+1)+1 \right]$ = 0

Mà:  $\frac{1}{4}{{(x+1)}^{2}}+\frac{1}{2}(x+1)+1$   > 0  với mọi x

Suy ra: $\frac{1}{2}(x+1)-1$ = 0

$\frac{1}{2}x+\frac{1}{2}-1=0$

$\frac{1}{2}x=\frac{1}{2}$

x = 1

vậy x =  1