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Question Number 191220 by anr0h3 last updated on 21/Apr/23

$$x^6-x^3=2solveforx$$$$$$$$m=x^3$$$$m^2={\left(x^3\right)}^2$$$$m^2-m=2$$$$m^2-m-2=0$$$$\alpha +\beta =1\to \alpha =\frac{1}{2}+\mu and\beta =\frac{1}{2}-\mu$$$$\alpha \cdot \beta =(\frac{1}{2}+\mu )\cdot (\frac{1}{2}-\mu )=-2$$$$\alpha \cdot \beta =(\frac{1}{2}-\mu )\cdot (\frac{1}{2}+\mu )=-2$$$$\frac{1}{4}-{\mu }^2=-2$$$${\mu }^2=2+\frac{1}{4}=\frac{9}{4}$$$$\mu =\sqrt{\frac{9}{4}}=\frac{3}{2}$$$$\alpha =\frac{1}{2}+\frac{3}{2}=\frac{-1+3}{2}$$$$\beta =\frac{1}{2}-\frac{3}{2}=\frac{-1-3}{2}$$$$(m+\frac{1+3}{2})(m+\frac{1-3}{2})=0$$$$m_1=\frac{1+3}{2}$$$$m_2=-\frac{1-3}{2}$$$$m_{1,2}=\frac{1\pm 3}{2}$$$$m=x^3\to x={\left(\frac{1\pm 3}{2}\right)}^{1/3}$$$$$$

Commented by Tinku Tara last updated on 21/Apr/23

$$(-\frac{1}{2}+\mu )(-\frac{1}{2}-\mu )=-2$$$$\Rightarrow \frac{1}{4}-{\mu }^2=-2$$$$\mathrm{You}\mathrm{made}\mathrm{a}\mathrm{mistake}\mathrm{in}\mathrm{sign}\mathrm{at}$$$$\mathrm{this}\mathrm{step}$$$$$$

Commented by anr0h3 last updated on 21/Apr/23

$$x^6-x^3=2solveforx$$$$$$$$m=x^3$$$$m^2={\left(x^3\right)}^2$$$$m^2-m=2$$$$m^2-m-2=0$$$$\alpha +\beta =1\to \alpha =\frac{1}{2}+\mu and\beta =\frac{1}{2}-\mu$$$$\alpha \cdot \beta =(\frac{1}{2}+\mu )\cdot (\frac{1}{2}-\mu )=-2$$$$\alpha \cdot \beta =(\frac{1}{2}-\mu )\cdot (\frac{1}{2}+\mu )=-2$$$$\frac{1}{4}-{\mu }^2=-2$$$${\mu }^2=2+\frac{1}{4}=\frac{9}{4}$$$$\mu =\sqrt{\frac{9}{4}}=\frac{3}{2}$$$$\alpha =\frac{1}{2}+\frac{3}{2}=\frac{1+3}{2}$$$$\beta =\frac{1}{2}-\frac{3}{2}=\frac{1-3}{2}$$$$(m+\frac{1+3}{2})(m+\frac{1-3}{2})=0$$$$m_1=\frac{1+3}{2}$$$$m_2=\frac{1-3}{2}$$$$m_{1,2}=\frac{1\pm 3}{2}$$$$m=x^3\to x={\left(\frac{1\pm 3}{2}\right)}^{1/3}$$$$$$

Commented by Tinku Tara last updated on 21/Apr/23

$$\alpha +\beta =1(not-1)$$

Commented by anr0h3 last updated on 21/Apr/23

$$x^6-x^3=2solveforx$$$$$$$$m=x^3$$$$m^2={\left(x^3\right)}^2$$$$m^2-m=2$$$$m^2-m-2=0$$$$\alpha +\beta =1\to \alpha =\frac{1}{2}+\mu and\beta =\frac{1}{2}-\mu$$$$\alpha \cdot \beta =(\frac{1}{2}+\mu )\cdot (\frac{1}{2}-\mu )=-2$$$$\alpha \cdot \beta =(\frac{1}{2}-\mu )\cdot (\frac{1}{2}+\mu )=-2$$$$\frac{1}{4}-{\mu }^2=-2$$$${\mu }^2=2+\frac{1}{4}=\frac{9}{4}$$$$\mu =\sqrt{\frac{9}{4}}=\frac{3}{2}$$$$\alpha =\frac{1}{2}+\frac{3}{2}=\frac{1+3}{2}$$$$\beta =\frac{1}{2}-\frac{3}{2}=\frac{1-3}{2}$$$$(m+\frac{1+3}{2})(m+\frac{1-3}{2})=0$$$$m_1=\frac{1+3}{2}$$$$m_2=\frac{1-3}{2}$$$$m_{1,2}=\frac{1\pm 3}{2}$$$$m=x^3\to x={\left(\frac{1\pm 3}{2}\right)}^{1/3}$$$$$$

Commented by anr0h3 last updated on 21/Apr/23

thnks, I was clearing my doubts whit the PO shen technique, and now I'm learning about the complex solution in this problem, isn't sarcasm, again thnks

Answered by Frix last updated on 21/Apr/23

$$\mathrm{What}\mathrm{have}\mathrm{you}\mathrm{done}???$$$$m^2-m-2=0$$$$\mathrm{Use}\mathrm{formula}:X^2+pX+q=0\Rightarrow X=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}$$$$\mathrm{Here}p=-1\wedge q=-2$$$$m=\frac{1}{2}\pm \sqrt{\frac{1}{4}+2}=\frac{1}{2}\pm \frac{3}{2}$$$$m_1=-1$$$$m_2=2$$$$x_{1,3,5}^3=-1$$$$x_{2,4,6}^3=2$$$$x_1=-1x_3=-\omega x_5=-{\omega }^2$$$$x_2=\sqrt[3]{2}{x}_4=\sqrt[3]{2}\omega x_5=\sqrt[3]{2}{\omega }^2$$$$[\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}]$$

Commented by anr0h3 last updated on 21/Apr/23

sorry, I have a mistake sign but the corrections have done, thnks

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