Question-224150

Question Number 224150 by behi834171 last updated on 23/Aug/25

Commented by behi834171 last updated on 23/Aug/25

$$\boldsymbol{{x}};\:\boldsymbol{{in}}\:\boldsymbol{{terms}}\:\boldsymbol{{of}}:\:\left(\boldsymbol{{a}}\:\boldsymbol{{and}}\:\boldsymbol{{b}}\right)\in\boldsymbol{{R}} \\ $$

Commented by Ghisom_ last updated on 23/Aug/25

p=−((4a^4 −3b^2 )/3)∧q=−((128a^6 −144a^2 b^2 +27)/(216)) △=(p^3 /(27))+(q^2 /4) △≥0 ⇒ x_1 ∈R∧x_(2, 3) ∉R u=−(q/2)+(√((p^3 /(27))+(q^2 /4)))∧v=−(q/2)−(√((p^3 /(27))+(q^2 /4))) ω=−(1/2)+((√3)/2)i x_1 =((2a^2 )/3)+u^(1/3) +v^(1/3) x_2 =((2a^2 )/3)+ωu^(1/3) +ω^2 v^(1/3) x_3 =((2a^2 )/3)+ω^2 u^(1/3) +ωv^(1/3) ...now unsert u, v, p, q △<0 ⇒ x_k ∈R k=0, 1, 2 x_k =(2/3)(a^2 +((√(−3p)) sin ((2kπ+arcsin ((3(√3) q)/(2(√(−p^3 )))))/3))) ...now insert p, q, k there is no “nice” solution

$${p}=−\frac{\mathrm{4}{a}^{\mathrm{4}} −\mathrm{3}{b}^{\mathrm{2}} }{\mathrm{3}}\wedge{q}=−\frac{\mathrm{128}{a}^{\mathrm{6}} −\mathrm{144}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{27}}{\mathrm{216}} \\ $$$$\bigtriangleup=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\bigtriangleup\geqslant\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1}} \in\mathbb{R}\wedge{x}_{\mathrm{2},\:\mathrm{3}} \notin\mathbb{R} \\ $$$${u}=−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}\wedge{v}=−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}}+{u}^{\mathrm{1}/\mathrm{3}} +{v}^{\mathrm{1}/\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}}+\omega{u}^{\mathrm{1}/\mathrm{3}} +\omega^{\mathrm{2}} {v}^{\mathrm{1}/\mathrm{3}} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}}+\omega^{\mathrm{2}} {u}^{\mathrm{1}/\mathrm{3}} +\omega{v}^{\mathrm{1}/\mathrm{3}} \\ $$$$…\mathrm{now}\:\mathrm{unsert}\:{u},\:{v},\:{p},\:{q} \\ $$$$ \\ $$$$\bigtriangleup<\mathrm{0}\:\Rightarrow\:{x}_{{k}} \in\mathbb{R} \\ $$$${k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$${x}_{{k}} =\frac{\mathrm{2}}{\mathrm{3}}\left({a}^{\mathrm{2}} +\left(\sqrt{−\mathrm{3}{p}}\:\mathrm{sin}\:\frac{\mathrm{2}{k}\pi+\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}\:{q}}{\mathrm{2}\sqrt{−{p}^{\mathrm{3}} }}}{\mathrm{3}}\right)\right) \\ $$$$…\mathrm{now}\:\mathrm{insert}\:{p},\:{q},\:{k} \\ $$$$ \\ $$$$\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution} \\ $$

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