#### Answer

$(0,1)$ and $(-1, 0)$ are on the graph

#### Work Step by Step

In order to determine if a point is on the graph of the equation, we have to substitute into the equation: $y^3=x+1$
For the point of $(1,2)$ substitute $x=1$ and $y=2$ into the equation
$2^3=1+1\\
8\ne2$
The equation is not satisfied, therefore the point is not on the graph.
For $(0,1)$:
$x=0$ and $y=1$
$1^3=0+1\\
1=1$
The equation is satisfied, therefore the point is on the graph.
For $(-1,0)$:
$x=-1$ $y=0$
$0^3=-1+1\\
0=0$
The equation is satisfied, therefore the point is on the graph.