Answer
$\bf \color {blue}{(0,0),\pm(1,1),\pm(2,8)}$
Work Step by Step
$$\bf{y=x^3}$$
$\bf \text{Let's use positive and negative perfect cubes for }y$
if $y=\pm0$:
$0=x^3$
$0=x$
$\bf \color {blue}{(0,0)}$
if $y=\pm1$:
$1=x^3$ or $-1=x^3$
$1=x$ or $-1=x$
$\bf \color {blue}{(1,1),(-1,-1)}$
if $y=\pm8$:
$8=x^3$ or $-8=x^3$
$2=x$ or $-2=x$
$\bf \color {blue}{(2,8),(-2,-8)}$
So some of the points are:
$\bf \color {blue}{(0,0),\pm(1,1),\pm(2,8)}$
Which we can graph as:
