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Question Number 215708 by essaad last updated on 15/Jan/25

Answered by A5T last updated on 15/Jan/25

$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\Rightarrow {\mathrm{n}}^3+{\mathrm{y}}^3=0\Rightarrow (\mathrm{n}+\mathrm{y})({\mathrm{n}}^2+{\mathrm{y}}^2-\mathrm{ny})=0...\left(\mathrm{iii}\right)$$$$\left(\mathrm{i}\right)-\left(\mathrm{ii}\right)\Rightarrow {4\mathrm{n}}^3-{4\mathrm{y}}^3-8\mathrm{n}+8\mathrm{y}=0$$$$\Rightarrow {\mathrm{n}}^3-{\mathrm{y}}^3-2\mathrm{n}+2\mathrm{y}=0$$$$(\mathrm{n}-\mathrm{y})({\mathrm{n}}^2+{\mathrm{y}}^2+\mathrm{ny}-2)=0...\left(\mathrm{iv}\right)$$$$\left(\mathrm{iii}\right)\mathrm{\&}\left(\mathrm{iv}\right)\mathrm{must}\mathrm{be}\mathrm{true}\mathrm{implies}$$$$\mathrm{Case}\mathrm{I}:\mathrm{n}+\mathrm{y}=0\mathrm{and}{\mathrm{n}}^2+{\mathrm{y}}^2+\mathrm{ny}-2=0$$$$\Rightarrow {\mathrm{y}}^2=2\Rightarrow \mathrm{y}=\underset{-}{+}\sqrt{2}$$$$\Rightarrow (\mathrm{n},\mathrm{y})=(\sqrt{2},-\sqrt{2});(-\sqrt{2},\sqrt{2})$$$$\mathrm{Case}\mathrm{II}:\mathrm{n}+\mathrm{y}=0\mathrm{and}\mathrm{n}-\mathrm{y}=0\Rightarrow (\mathrm{n},\mathrm{y})=(0,0)$$$$\mathrm{Case}\mathrm{III}:{\mathrm{n}}^2+{\mathrm{y}}^2-\mathrm{ny}=0\mathrm{and}\mathrm{n}-\mathrm{y}=0\Rightarrow (\mathrm{n},\mathrm{y})=(0,0)$$$$\mathrm{Case}\mathrm{IV}:{\mathrm{n}}^2+{\mathrm{y}}^2-\mathrm{ny}=0\mathrm{and}{\mathrm{n}}^2+{\mathrm{y}}^2+\mathrm{ny}-2=0$$$$\Rightarrow {(\mathrm{n}+\mathrm{y})}^2=3\mathrm{ny}\mathrm{and}{(\mathrm{n}+\mathrm{y})}^2=\mathrm{ny}+2$$$$\Rightarrow \mathrm{ny}=1\Rightarrow (\mathrm{n}+\mathrm{y})=\underset{-}{+}\sqrt{3}$$$$\mathrm{n},\mathrm{y}\mathrm{satisfy}{\mathrm{x}}^2\stackrel{-}{+}\sqrt{3}\mathrm{x}+1=0$$$$\Rightarrow \mathrm{n},\mathrm{y}=\frac{\left(\underset{-}{+}\right)\sqrt{3}\underset{-}{+}\sqrt{3-4\left(1\right)}}{2}=\frac{\left(\underset{-}{+}\right)\sqrt{3}\underset{-}{+}\mathrm{i}}{2}$$$$(\mathrm{n},\mathrm{y})$$$$=(\frac{\sqrt{3}+\mathrm{i}}{2},\frac{\sqrt{3}-\mathrm{i}}{2});(\frac{-\sqrt{3}+\mathrm{i}}{2},\frac{-\sqrt{3}-\mathrm{i}}{2});(\sqrt{2},-\sqrt{2});(0,0)$$$$\mathrm{upto}\mathrm{permutation}\mathrm{since}\mathrm{equation}\mathrm{is}\mathrm{symmetric}.$$

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