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Question Number 144768 by ajfour last updated on 04/Jul/21
x^3 −x−c=0 let x=(√(t+p))+(√(t+q)) (t+p)(√(t+p))+(t+q)(√(t+q)) +3(t+p)(√(t+q))+3(t+q)(√(t+p)) −(√(t+p))−(√(t+q))−c=0 let (t+q)+3(t+p)−1=0 ⇒ 4t=1−(p+q) ⇒ 4t+4p=3p−q+1 4t+4q=1−(p−3q) (3p−q+1)^(3/2) +3(1−p+3q)(√(3p−q+1)) −4(√(3p−q+1))−8c=0 2x=(√(3p−q+1))+(√(3q−p+1)) let (√(3p−q+1))=−1 ⇒ q=3p ⇒ −1−3(1−p+3q)+4−8c=0 ⇒ p=−(c/3) 2x=−1+(√(1−((8c)/3))) x=−(1/2)+(√((1/4)−((2c)/3))) for c=(1/4) x=−(1/2)+(1/(2(√3)))
$$\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\sqrt{{t}+{p}}+\sqrt{{t}+{q}} \\ $$$$\left({t}+{p}\right)\sqrt{{t}+{p}}+\left({t}+{q}\right)\sqrt{{t}+{q}} \\ $$$$+\mathrm{3}\left({t}+{p}\right)\sqrt{{t}+{q}}+\mathrm{3}\left({t}+{q}\right)\sqrt{{t}+{p}} \\ $$$$−\sqrt{{t}+{p}}−\sqrt{{t}+{q}}−{c}=\mathrm{0} \\ $$$${let}\:\:\left({t}+{q}\right)+\mathrm{3}\left({t}+{p}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{4}{t}=\mathrm{1}−\left({p}+{q}\right) \\ $$$$\Rightarrow\:\:\mathrm{4}{t}+\mathrm{4}{p}=\mathrm{3}{p}−{q}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\mathrm{4}{t}+\mathrm{4}{q}=\mathrm{1}−\left({p}−\mathrm{3}{q}\right) \\ $$$$\left(\mathrm{3}{p}−{q}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} +\mathrm{3}\left(\mathrm{1}−{p}+\mathrm{3}{q}\right)\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}} \\ $$$$−\mathrm{4}\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}}−\mathrm{8}{c}=\mathrm{0} \\ $$$$\mathrm{2}{x}=\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}}+\sqrt{\mathrm{3}{q}−{p}+\mathrm{1}} \\ $$$${let}\:\:\sqrt{\mathrm{3}{p}−{q}+\mathrm{1}}=−\mathrm{1} \\ $$$$\Rightarrow\:\:{q}=\mathrm{3}{p} \\ $$$$\Rightarrow\:−\mathrm{1}−\mathrm{3}\left(\mathrm{1}−{p}+\mathrm{3}{q}\right)+\mathrm{4}−\mathrm{8}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{p}=−\frac{{c}}{\mathrm{3}} \\ $$$$\mathrm{2}{x}=−\mathrm{1}+\sqrt{\mathrm{1}−\frac{\mathrm{8}{c}}{\mathrm{3}}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}{c}}{\mathrm{3}}} \\ $$$${for}\:\:{c}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

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