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Question Number 96584 by Jidda28 last updated on 04/Jun/20

Commented by mr W last updated on 03/Jun/20

$$whenpostinganimageyoushould$$$$croptheimageproperlyordiasable$$$$crop.$$

Commented by prakash jain last updated on 04/Jun/20

$$\mathrm{Hi}$$$$\mathrm{Please}\mathrm{move}/\mathrm{resize}\mathrm{black}\mathrm{rectangle}$$$$\mathrm{to}\mathrm{select}\mathrm{appropriate}\mathrm{crop}\mathrm{area}.$$$$\mathrm{If}\mathrm{you}\mathrm{wish}\mathrm{to}\mathrm{upload}\mathrm{entire}\mathrm{image}.$$$$\mathrm{Just}\mathrm{disable}\mathrm{crop}.$$

Commented by Jidda28 last updated on 04/Jun/20

$$\mathrm{ok}\mathrm{sir}\mathrm{i}\mathrm{edited}\mathrm{the}\mathrm{post}.$$

Commented by Jidda28 last updated on 04/Jun/20

$$\mathrm{ok}\mathrm{sir}\mathrm{i}\mathrm{edited}\mathrm{the}\mathrm{post}.$$

Answered by prakash jain last updated on 04/Jun/20

$$y^3+{xy}^2=2$$$$x=\frac{2-y^3}{y^2}$$$${\left(\frac{2-y^3}{y^2}\right)}^3+9y{\left(\frac{2-y^3}{y^2}\right)}^2=10$$$${(\frac{2}{y^2}-y)}^3+9y{(\frac{2}{y^2}-y)}^2=10$$$$-y^3+6-\frac{12}{y^3}+\frac{8}{y^6}+\frac{36}{y^3}+9y^3-36=10$$$$8y^3-30+\frac{24}{y^3}+\frac{8}{y^6}=10$$$$8y^3-40+\frac{24}{y^3}+\frac{8}{y^6}=10$$$$y^3-5+\frac{3}{y^3}+\frac{1}{y^6}=0$$$$y^3=u$$$$u-5+\frac{3}{u}+\frac{1}{u^2}=0$$$$u^3-5u^2+3u+1=0$$$$u^3-u^2-4u^2+4u-u+1=0$$$$u^2(u-1)-4u(u-1)-(u-1)=9$$$$(u-1)(u^2-4u-1)=0\left(\mathrm{I}\right)$$$$u=1$$$$u=\frac{4\pm \sqrt{20}}{2}=2\pm \sqrt{5}$$$$\mathrm{Considering}\mathrm{only}\mathrm{real}\mathrm{roots}$$$$u=y^3=1\Rightarrow y=1$$$$u=y^3=2\pm \sqrt{5}$$$$={\left(\frac{1}{2}(1\pm \sqrt{5})\right)}^3\Rightarrow y=\frac{1\pm \sqrt{5}}{2}$$$$x=\frac{2-y^3}{y^2}$$$$y=1\Rightarrow x=1$$$$y=\frac{1\pm \sqrt{5}}{2}\Rightarrow x=\frac{2-(2\pm \sqrt{5})}{\frac{1}{4}{(1\pm \sqrt{5})}^2}=\frac{\mp 4\sqrt{5}}{6\pm 2\sqrt{5}}$$$$x=\frac{\mp 4\sqrt{5}(6\mp 2\sqrt{5})}{6^2-20}=\frac{1}{2}(5\mp 3\sqrt{5})$$$$\mathrm{If}\mathrm{you}\mathrm{need}\mathrm{you}\mathrm{can}\mathrm{calculate}$$$$\mathrm{complex}\mathrm{roots}\mathrm{from}\left(\mathrm{I}\right)$$

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