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Question Number 18392 by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17

Commented by b.e.h.i.8.3.417@gmail.com last updated on 20/Jul/17

$$solveforx,y,z.$$

Answered by mrW1 last updated on 20/Jul/17

$$\left(\mathrm{i}\right)-\left(\mathrm{ii}\right):$$$${\mathrm{x}}^2-{\mathrm{y}}^2=\mathrm{y}-\mathrm{x}$$$$(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})=-(\mathrm{x}-\mathrm{y})$$$$\mathrm{x}-\mathrm{y}=0\mathrm{or}\mathrm{x}+\mathrm{y}=-1(\mathrm{i}.\mathrm{e}.{\mathrm{z}}^2=-1\Rightarrow \mathrm{no}\mathrm{real}\mathrm{roots})$$$$$$$$\mathrm{with}\mathrm{x}=\mathrm{y}$$$${\mathrm{y}}^2=\mathrm{z}+\mathrm{y}$$$${\mathrm{z}}^2=2\mathrm{y}$$$${({\mathrm{y}}^2-\mathrm{y})}^2=2\mathrm{y}$$$${\mathrm{y}}^2{(\mathrm{y}-1)}^2=2\mathrm{y}$$$$\Rightarrow \mathrm{y}=0\mathrm{or}$$$$\Rightarrow \mathrm{y}{(\mathrm{y}-1)}^2=2$$$$\Rightarrow \mathrm{y}({\mathrm{y}}^2-2\mathrm{y}+1)=2$$$$\Rightarrow {\mathrm{y}}^3-{2\mathrm{y}}^2+\mathrm{y}-2=0$$$$\Rightarrow (\mathrm{y}-2)({\mathrm{y}}^2+1)=0$$$$\Rightarrow \mathrm{y}=2\mathrm{or}\mathrm{y}=\pm \mathrm{i}$$$$\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=0$$$$\Rightarrow \mathrm{x}=\mathrm{y}=2\Rightarrow \mathrm{z}=2$$$$\Rightarrow \mathrm{x}=\mathrm{y}=\pm \mathrm{i}\Rightarrow \mathrm{z}=-1\mp \mathrm{i}$$$$$$$$\mathrm{with}\mathrm{x}+\mathrm{y}=-1\mathrm{or}\mathrm{x}=-1-\mathrm{y}$$$${\mathrm{z}}^2=\mathrm{x}+\mathrm{y}=-1$$$$\Rightarrow \mathrm{z}=\mp \mathrm{i}$$$${\mathrm{y}}^2=\mathrm{z}-1-\mathrm{y}$$$${\mathrm{y}}^2+\mathrm{y}+1\pm \mathrm{i}=0$$$$\Rightarrow \mathrm{y}=\frac{-1\pm \sqrt{1-4(1\pm \mathrm{i})}}{2}=\frac{-1\pm \sqrt{-3\mp \mathrm{i}}}{2}$$$$\Rightarrow \mathrm{x}=-1-\mathrm{y}=\frac{-1\mp \sqrt{-3\mp \mathrm{i}}}{2}$$$$$$$$\mathrm{in}\mathrm{total}$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(0,0,0)$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(2,2,2)$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(\mathrm{i},\mathrm{i},-1-\mathrm{i})$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(-\mathrm{i},-\mathrm{i},-1+\mathrm{i})$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(\frac{-1-\sqrt{-3-\mathrm{i}}}{2},\frac{-1+\sqrt{-3-\mathrm{i}}}{2},-\mathrm{i})$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(\frac{-1+\sqrt{-3-\mathrm{i}}}{2},\frac{-1-\sqrt{-3-\mathrm{i}}}{2},-\mathrm{i})$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(\frac{-1-\sqrt{-3+\mathrm{i}}}{2},\frac{-1+\sqrt{-3+\mathrm{i}}}{2},\mathrm{i})$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(\frac{-1+\sqrt{-3+\mathrm{i}}}{2},\frac{-1-\sqrt{-3+\mathrm{i}}}{2},\mathrm{i})$$

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