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Question Number 174059 by mr W last updated on 23/Jul/22

$$if{x}^3+y^3+3xy=1,thenx+y=?$$

Commented by infinityaction last updated on 23/Jul/22

$$y=0andx=1$$$$thenx+y=1$$$$orx=-1andy=-1$$$$thenx+y=-2$$

Commented by Tawa11 last updated on 24/Jul/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by dragan91 last updated on 24/Jul/22

$$(\mathrm{x}+\mathrm{y})({\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2)=1-3\mathrm{xy}$$$$(\mathrm{x}+\mathrm{y})({(\mathrm{x}+\mathrm{y})}^2-3\mathrm{xy})=1-3\mathrm{xy}$$$${(\mathrm{x}+\mathrm{y})}^3-1-3\mathrm{xy}(\mathrm{x}+\mathrm{y})+3\mathrm{xy}=0$$$$(\mathrm{x}+\mathrm{y}-1)({(\mathrm{x}+\mathrm{y})}^2+\mathrm{x}+\mathrm{y}+1)-3\mathrm{xy}(\mathrm{x}+\mathrm{y}-1)=0$$$$(\mathrm{x}+\mathrm{y}-1)({\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{y}}^2+\mathrm{x}+\mathrm{y}+1-3\mathrm{xy})=0$$$$1)\mathrm{x}+\mathrm{y}=1$$$$2){\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2+\mathrm{x}+\mathrm{y}+1=0$$$${\mathrm{x}}^2+\mathrm{x}(1-\mathrm{y})+{\mathrm{y}}^2+\mathrm{y}+1=0$$$${\mathrm{x}}_{1,2}=\frac{\mathrm{y}-1\pm \sqrt{1-2\mathrm{y}+{\mathrm{y}}^2-{4\mathrm{y}}^2-4\mathrm{y}-4}}{2}$$$$=\frac{\mathrm{y}-1}{2}\pm \frac{\sqrt{-3({\mathrm{y}}^2+2\mathrm{y}+1)}}{2}$$$$=\frac{\mathrm{y}-1}{2}\pm \frac{(\mathrm{y}+1)\mathrm{i}\sqrt{3}}{2}$$$$\Rightarrow \mathrm{x}+\mathrm{y}=\frac{3\mathrm{y}-1}{2}\pm \frac{(\mathrm{y}+1)\mathrm{i}\sqrt{3}}{2},\mathrm{y}\in \mathrm{R}$$$$$$

Answered by behi834171 last updated on 24/Jul/22

$$x+y=p,xy=q$$$$(\mathit{x}+\mathit{y})[{(\mathit{x}+\mathit{y})}^2-3\mathit{x}\mathit{y}]+3\mathit{x}\mathit{y}=1$$$$\Rightarrow \mathit{p}({\mathit{p}}^2-3\mathit{q})+3\mathit{q}=1\Rightarrow {\mathit{p}}^3-3\mathit{p}\mathit{q}+3\mathit{q}-1=0$$$$(\mathit{p}-1)({\mathit{p}}^2+\mathit{p}-3\mathit{q}+1)=0$$$$\Rightarrow \mathit{p}=1,\mathit{p}=\frac{-1\pm \sqrt{12\mathit{q}-3}}{2}$$$$\Rightarrow \mathit{x}+\mathit{y}=\mathit{p}=1.◼[onlypossibility,ithink]$$$$[12\mathit{q}-3=0\Rightarrow \mathit{q}=\frac{1}{4},\mathit{p}=-\frac{1}{2}$$$$\Rightarrow {\mathit{z}}^2+\frac{\mathit{z}}{2}+\frac{1}{4}=0\Rightarrow 4z^2+2z+1=0$$$$z=\frac{-2\pm \sqrt{4-16}}{8}=\frac{-2\pm 2\sqrt{3}\mathit{i}}{8}=-(\frac{1}{4}\mp \mathit{i}.\frac{\sqrt{3}}{4})$$$$\Rightarrow \{\begin{array}{l}\mathit{x}=-(\frac{1}{4}+\mathit{i}.\frac{\sqrt{3}}{4})\\ \mathit{y}=-(\frac{1}{4}+\mathit{i}.\frac{\sqrt{3}}{4})\end{array}\Rightarrow \mathit{x}+\mathit{y}\stackrel{?}{=}-\frac{1}{2}$$$${\mathit{x}}^3+{\mathit{y}}^3=(\mathit{x}+\mathit{y})[{(\mathit{x}+\mathit{y})}^2-3\mathit{x}\mathit{y}]=$$$$=(-\frac{1}{2})[\frac{1}{4}-\frac{3}{4}]=\frac{1}{4}$$$$\Rightarrow {\mathit{x}}^3+{\mathit{y}}^3+3\mathit{x}\mathit{y}=\frac{1}{4}+3\times \frac{1}{4}=1\mathit{o}\mathit{k}$$$$\Rightarrow \mathit{x}+\mathit{y}=-\frac{1}{2}\left[\mathit{a}\mathit{n}\mathit{o}\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{p}\mathit{o}\mathit{s}\mathit{s}\mathit{i}\mathit{b}\mathit{i}\mathit{l}\mathit{i}\mathit{t}\mathit{y}\right]]$$

Answered by MJS_new last updated on 24/Jul/22

$$x^3+y^3+3xy-1=0$$$$(x+y-1)(x^2-xy+y^2+x+y+1)=0$$$$\left(1\right)x+y=1$$$$\left(2\right)y=\frac{x-1}{2}\pm \frac{\sqrt{3}(x+1)}{2}\mathrm{i}$$$$\mathrm{if}x,y\in \mathbb{R}\Rightarrow x=-1\wedge y=-1\Rightarrow x+y=-2$$$$\mathrm{if}x,y\in \mathbb{C}\Rightarrow x+y=\frac{3x-1}{2}\pm \frac{\sqrt{3}(x+1)}{2}\mathrm{i}\wedge x\in \mathbb{C}$$

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