Answer
The solution is $\left\{ \left( 0,0 \right),\left( -1,1 \right) \right\}$
Work Step by Step
The given equations are,
$\begin{align}
& {{x}^{3}}+y=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& {{x}^{2}}-y=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\
\end{align}$
For solving the equation use the addition method:
Add equation (I) and (II) to eliminate y:
$\begin{align}
& {{x}^{3}}+y=0 \\
& \underline{{{x}^{2}}-y=0} \\
& {{x}^{3}}+{{x}^{2}}=0 \\
\end{align}$
${{x}^{2}}\left( x+1 \right)=0$ (III)
${{x}^{2}}=0$ or $ x+1=0$
$ x=0$ or $ x=-1$
If $ x=0$ then substitute $ x=0$ in equation (I):
$\begin{align}
& {{\left( 0 \right)}^{3}}+y=0 \\
& y=0
\end{align}$
If $ x=-1$ then substitute $ x=-1$ in equation (I):
$\begin{align}
& {{\left( -1 \right)}^{3}}+y=0 \\
& y=1
\end{align}$
Therefore, the solution set is $\left\{ \left( 0,0 \right),\left( -1,1 \right) \right\}$
Hence, the solution set of the system is, $\left\{ \left( 0,0 \right),\left( -1,1 \right) \right\}$.
