Question Number 64066 by mathmax by abdo last updated on 12/Jul/19
$${let}\:\alpha\:,\beta\:{and}\:\lambda\:{the}\:{roots}\:{of}\:{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}\:=\mathrm{0}\:{find}\:{the}\:{value}\:{of} \\ $$$${A}\:=\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \:{and}\:\:{B}\:=\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \:+\lambda^{\mathrm{3}} \:. \\ $$
Answered by MJS last updated on 12/Jul/19
$$\mathrm{the}\:\mathrm{solutions}\:\mathrm{are} \\ $$$$\alpha={u}+{v} \\ $$$$\beta=\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$$\lambda=\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${A}=\mathrm{6}{uv} \\ $$$${B}=\mathrm{3}\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right) \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right.}}{\mathrm{18}}} \\ $$$${v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right.}}{\mathrm{18}}} \\ $$$${A}=−\mathrm{2}{p} \\ $$$${B}=−\mathrm{3}{q} \\ $$$${p}=\mathrm{2} \\ $$$${q}=−\mathrm{1} \\ $$$${A}=−\mathrm{4} \\ $$$${B}=\mathrm{3} \\ $$
Commented by mathmax by abdo last updated on 12/Jul/19
$${thank}\:{you}\:{sir}\:{mjs} \\ $$