$$ \begin{cases} y+1 = 4(x+1)^3\\ x+1 = 4(y+1)^3. \end{cases} $$So $y+1=4^4(y+1)^9 \to y=-1$ or $2^8(y+1)^8 =1 \to 2(y+1)=\pm 1 \to y=-\frac{1}{2},-\frac{3}{2}$ Answer: $(-1;-1),(-\frac{1}{2},-\frac{1}{2});(-\frac{3}{2},-\frac{3}{2})$

Real: $(-1,-1)$, $\left(-\frac{3}{2},-\frac{3}{2}\right)$, $\left(-\frac{1}{2},-\frac{1}{2}\right)$ Complex: $\left(-1\pm\frac{i}{2},-1\mp\frac{i}{2}\right)$, $\left(-1+\frac{\sqrt{2}}{4}\pm\frac{\sqrt{2}}{4}i,-1-\frac{\sqrt{2}}{4}\pm\frac{\sqrt{2}}{4}i\right)$, $\left(-1-\frac{\sqrt{2}}{4}\pm\frac{\sqrt{2}}{4}i,-1+\frac{\sqrt{2}}{4}\pm\frac{\sqrt{2}}{4}i\right)$

How did you do @rchokler to find the Complex solutions??

LittleKesha wrote: How did you do @rchokler to find the Complex solutions?? roots of unity

Wolfram Alpha