Given, $z^3+\bar{z}=0$
Let $\quad z=x+i y$
$$
\begin{aligned}
& \therefore \quad(x+i y)^3+x-i y=0 \\
& \Rightarrow x^3+(i y)^3+3 x^2 i y+3 x(i y)^2+x-i y=0 \\
& \Rightarrow x^3-y^3 i+3 x^2 y i-3 x y^2+x-i y=0 \\
& \Rightarrow\left(x^3-3 x y^2+x\right)+\left(-y^3+3 x^2 y-y\right) i=0
\end{aligned}
$$
On equating real and imaginary parts, we get
$$
\begin{aligned}
& x^3-3 x y^2+x=0 \\
& \text { or } \quad-y^3+3 x^2 y-y=0 \\
& \Rightarrow \quad x\left(x^2-3 y^2+1\right)=0 \\
& \text { or } \quad-y\left(y^2-3 x^2+1\right)=0 \\
& \Rightarrow \quad x=0 \text { and } x^2-3 y^2+1=0 \\
& \text { or } y=0 \text { and } y^2-3 x^2+1=0 \\
& \text { Now, } \\
& x^3-3 y^2+1=y^2-3 x^2+1 \\
& 4 x^2=4 y^2 \\
& x= \pm y \\
& y^2-3 y^2+1=0 \\
& \Rightarrow \quad 2 y^2=1 \Rightarrow y= \pm \frac{1}{2} \\
& \Rightarrow \quad x= \pm \frac{1}{2} \\
&
\end{aligned}
$$
Hence, their solutions are $(0,0)$ and $\left( \pm \frac{1}{2}, \pm \frac{1}{2}\right)$
Hence, number of solutions is 5 .