#### Answer

$(0,0)$ and $(1,-1)$ 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: $y=x^3-2\sqrt{x}$
For the point of $(0,0)$ substitute $x=0$ and $y=0$ into the equation
$0=0^3-2\sqrt{0}\\
0=0$
The equation is satisfied, therefore the point is on the graph.
For $(1,1)$:
$x=1$ and $y=1$
$1=1^3-2\sqrt{1}\\
1\ne-1$
The equation is not satisfied, therefore the point is not on the graph.
For $(1,-1)$:
$x=1$ $y=-1$
$-1=1^3-2\sqrt{1}\\
-1=-1$
The equation is satisfied, therefore the point is on the graph.