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Question Number 64066 by mathmax by abdo last updated on 12/Jul/19

$$let\alpha ,\beta and\lambda therootsof{x}^3+2x-1=0findthevalueof$$$$A={\alpha }^2+{\beta }^2+{\lambda }^{2}andB={\alpha }^3+{\beta }^3+{\lambda }^3.$$

Answered by MJS last updated on 12/Jul/19

$$\mathrm{the}\mathrm{solutions}\mathrm{are}$$$$\alpha =u+v$$$$\beta =(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})v$$$$\lambda =(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})v$$$$A=6uv$$$$B=3(u^3+v^3)$$$$u=\sqrt[3]{-\frac{q}{2}+\frac{\sqrt{3(4p^3+27q^2}}{18}}$$$$v=\sqrt[3]{-\frac{q}{2}-\frac{\sqrt{3(4p^3+27q^2}}{18}}$$$$A=-2p$$$$B=-3q$$$$p=2$$$$q=-1$$$$A=-4$$$$B=3$$

Commented by mathmax by abdo last updated on 12/Jul/19

$$thankyousirmjs$$

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