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Question Number 209761 by pablo1234523 last updated on 20/Jul/24

$$x^3-9{xy}^2=28$$$$x^{2}y-y^3=15$$$$\mathrm{solve}\mathrm{for}x\mathrm{and}y.$$$$x,y\in \mathbb{R}$$

Answered by mr W last updated on 20/Jul/24

$$lety=kx$$$$x^3(1-9k^2)=28...\left(i\right)$$$$x^3(k-k^3)=15...\left(ii\right)$$$$\left(i\right)/\left(ii\right):$$$$\frac{1-9k^2}{k-k^3}=\frac{28}{15}$$$$28k^3-135k^2-28k+15=0$$$$(k-5)(4k-1)(7k+3)=0$$$$\Rightarrow k=-\frac{3}{7},\frac{1}{4},5$$$$\Rightarrow x=\sqrt[3]{\frac{28}{1-9k^2}}=-\frac{7}{2},4,-\frac{1}{2}$$$$\Rightarrow y=\frac{3}{2},1,-\frac{5}{2}$$

Commented by pablo1234523 last updated on 20/Jul/24

$$\mathrm{thanks}\mathrm{a}\mathrm{lot}\mathrm{sir};$$$$\mathrm{how}\mathrm{did}\mathrm{u}\mathrm{find}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{that}\mathrm{cubic}\mathrm{equation}?$$$$\mathrm{by}\mathrm{inspection}\mathrm{i}\mathrm{think}.$$$$\mathrm{my}\mathrm{actual}\mathrm{question}\mathrm{was}:-$$$$\mathrm{how}\mathrm{to}\mathrm{find}\sqrt[3]{28+45i\sqrt{3}}.$$

Commented by mr W last updated on 20/Jul/24

$$itriedwiththefactorsof15,and$$$$found5isaroot.therestisthen$$$$easy.$$

Commented by mr W last updated on 20/Jul/24

$$28+45\sqrt{3}i={\left(\sqrt{19}\right)}^3{e}^{i{\tan }^{-1}\frac{45\sqrt{3}}{28}}$$$$\sqrt[3]{28+45\sqrt{3}i}=\sqrt{19}{e}^{i(\frac{2k\pi }{3}+\frac{1}{3}{\tan }^{-1}\frac{45\sqrt{3}}{28})}withk=0,1,2$$

Commented by pablo1234523 last updated on 20/Jul/24

$$\mathrm{thanks}\mathrm{sir},$$$$\mathrm{infact}\sqrt[3]{28+45\mathrm{i}\sqrt{3}}=-\frac{7}{2}+\frac{3i\sqrt{3}}{2},4+i\sqrt{3},-\frac{1}{2}-\frac{5i\sqrt{3}}{2}$$$$\mathrm{is}\mathrm{there}\mathrm{method}\mathrm{to}\mathrm{find}\tan \left(\frac{1}{3}{\tan }^{-1}\frac{45\sqrt{3}}{28}\right)$$$$\mathrm{more}\mathrm{specifucally},\cos \left(\frac{1}{3}{\tan }^{-1}\frac{45\sqrt{3}}{28}\right)=\cos \left(\frac{1}{3}{\cos }^{-1}\frac{28}{19\sqrt{19}}\right)$$$$\mathrm{similarly}\mathrm{for}\sin (..)$$$$$$

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