Question Number 160923 by cortano last updated on 09/Dec/21
$$\:\:\begin{cases}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2021}} =\:\mathrm{z}}\\{\left(\mathrm{x}+\mathrm{z}\right)^{\mathrm{2021}} \:=\:\mathrm{y}}\\{\left(\mathrm{y}+\mathrm{z}\right)^{\mathrm{2021}} \:=\:\mathrm{x}}\end{cases} \\ $$$$\:\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=… \\ $$
Commented by Rasheed.Sindhi last updated on 09/Dec/21
$$\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$