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Question Number 139948 by qaz last updated on 02/May/21

$$If{x}^2+xy+y^2=0$$$$find::{\left(\frac{x}{x+y}\right)}^{2023}+{\left(\frac{y}{x+y}\right)}^{2023}$$

Answered by Rasheed.Sindhi last updated on 04/May/21

$$\frac{x}{y}+\frac{y}{x}=-1\Rightarrow \frac{y}{x}=-(1+\frac{x}{y})$$$$\frac{x}{y}=-(1+\frac{y}{x})$$$${\left(\frac{x}{x+y}\right)}^{2023}+{\left(\frac{y}{x+y}\right)}^{2023}$$$$={(1+\frac{y}{x})}^{-2023}+{(1+\frac{x}{y})}^{-2023}$$$$={(1-(1+\frac{x}{y}))}^{-2023}+(1-{(1+\frac{y}{x})}^{-2023}$$$$={(-\frac{x}{y})}^{-2023}+{(-\frac{y}{x})}^{-2023}$$$$={(-\frac{y}{x})}^{2023}+{(-\frac{x}{y})}^{2023}$$$$=-{\left(\frac{y}{x}\right)}^{2023}-{\left(\frac{x}{y}\right)}^{2023}$$$$\mathrm{Continue}$$

Answered by mathmax by abdo last updated on 02/May/21

$${\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2=0\mathrm{for}\mathrm{unknown}\mathrm{x}$$$$\mathrm{\Delta }={\mathrm{y}}^2-{4\mathrm{y}}^2=-{3\mathrm{y}}^2\Rightarrow {\mathrm{x}}_1=\frac{-\mathrm{y}+\mathrm{i}\sqrt{3}\mathrm{y}}{2}=\mathrm{y}{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}\mathrm{and}{\mathrm{x}}_2={\mathrm{ye}}^{-\frac{\mathrm{i}2\pi }{3}}$$$$\mathrm{for}\mathrm{x}={\mathrm{x}}_1\mathrm{we}\mathrm{get}\mathrm{\Phi }(\mathrm{x},\mathrm{y})={\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{y}}\right)}^{2023}+{\left(\frac{\mathrm{y}}{\mathrm{x}+\mathrm{y}}\right)}^{2023}$$$$={\left(\frac{{\mathrm{ye}}^{\frac{2\mathrm{i}\pi }{3}}}{\mathrm{y}{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}+\mathrm{y}}\right)}^{2023}+{\left(\frac{\mathrm{y}}{{\mathrm{ye}}^{\frac{\mathrm{i}2\pi }{3}}+\mathrm{y}}\right)}^{2023}={\left(\frac{{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}}{1+{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}}\right)}^{2023}+{\left(\frac{1}{1+{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}}\right)}^{2023}$$$$=\frac{{\left({\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}\right)}^{2023}+1}{{(1+{\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}})}^{2023}}\mathrm{we}\mathrm{have}1+\mathrm{j}+{\mathrm{j}}^2=0\Rightarrow 1+\mathrm{j}=-{\mathrm{j}}^2\Rightarrow {(1+\mathrm{j})}^{2023}=-{\mathrm{j}}^{4046}$$$$\Rightarrow \mathrm{\Phi }(\mathrm{x},\mathrm{y})=-\frac{1+{\mathrm{e}}^{\frac{4046\mathrm{i}\pi }{3}}}{{\left({\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}\right)}^{4046}}=-\frac{1+{\mathrm{e}}^{\frac{4046\mathrm{i}\pi }{3}}}{{\mathrm{e}}^{\frac{8092\mathrm{i}\pi }{3}}}$$$$=-{\mathrm{e}}^{\frac{8092\mathrm{i}\pi }{3}}(1+{\mathrm{e}}^{\frac{4046\mathrm{i}\pi }{3}})\mathrm{rest}\mathrm{simplification}\mathrm{of}\mathrm{\Phi }(\mathrm{x},\mathrm{y})....$$

Answered by mr W last updated on 02/May/21

$$lett=\frac{y}{x}$$$$x^2+{tx}^2+t^2{x}^2=0$$$$t^2+t+1=0\Rightarrow t+1=-t^2$$$$\frac{1}{t^2}+\frac{1}{t}+1=0$$$$\frac{1}{t}\ne 1$$$$(\frac{1}{t^2}+\frac{1}{t}+1)(\frac{1}{t}-1)=0$$$${\left(\frac{1}{t}\right)}^3-1=0$$$$\Rightarrow {\left(\frac{1}{t}\right)}^3=1$$$$$$$${\left(\frac{x}{x+y}\right)}^{2023}+{\left(\frac{y}{x+y}\right)}^{2023}$$$$={\left(\frac{1}{1+t}\right)}^{2023}+{\left(\frac{t}{1+t}\right)}^{2023}$$$$={(-\frac{1}{t^2})}^{2023}+{(-\frac{1}{t})}^{2023}$$$$=-{\left(\frac{1}{t^2}\right)}^{2023}-{\left(\frac{1}{t}\right)}^{2023}$$$$=-{\left(\frac{1}{t}\right)}^{3\times 1348}{\left(\frac{1}{t}\right)}^2-{\left(\frac{1}{t}\right)}^{3\times 674}\left(\frac{1}{t}\right)$$$$=-{\left(\frac{1}{t}\right)}^2-\left(\frac{1}{t}\right)$$$$=-[{\left(\frac{1}{t}\right)}^2+\left(\frac{1}{t}\right)+1]+1$$$$=-0+1$$$$=1$$

Commented by qaz last updated on 04/May/21

$$thankyouSirs.mr{W}^{\prime }swayiscorrect.$$

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