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x-x-3-x-9-8-1-3-2x-x-3-3x-1-3-

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Question Number 150527 by Jamshidbek last updated on 13/Aug/21
((x(x−3)(x−9)−8))^(1/3) =2x+((x^3 −3x))^(1/3)
$$\sqrt[{\mathrm{3}}]{\mathrm{x}\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}−\mathrm{9}\right)−\mathrm{8}}=\mathrm{2x}+\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} −\mathrm{3x}} \\ $$
Answered by MJS_new last updated on 13/Aug/21
trying something weird: assuming ((x^3 −3x))^(1/3) =y∈R ⇒ x^3 =3x+y^3 (1) x(x−3)(x−9)−8=x^3 −12x^2 +27x−8= =(3x+y^3 )−12x^2 +27x−8= =−12x^2 +30x+y^3 −8 now we have ((−12x^2 +30x+y^3 −8))^(1/3) =2x+y −12x^2 +30x+y^3 −8=(2x+y)^3 ⇔ x^3 +(3/2)x^2 (y+1)+(3/4)x(y^2 −5)+1=0 (2) (1) x^3 −3x−y^3 =0 (2) x^3 +(3/2)x^2 (y+1)+(3/4)x(y^2 −5)+1=0 now match them (a) (3/2)(y+1)=0 (b) (3/4)(y^2 −5)=−3 (c) −y^3 =1 ⇒ y=−1 now we have (1) = (2) x^3 −3x+1=0 ⇒ x_1 =−2cos (π/9) ≈−1.87938524 x_2 =2sin (π/(18)) ≈.347296355 x_3 =2cos ((2π)/9) ≈1.53208889 testing these with the rule for r∈R: ((−r))^(1/3) =−(r)^(1/3) ⇒ they all fit the given equation
$$\mathrm{trying}\:\mathrm{something}\:\mathrm{weird}: \\ $$$$\mathrm{assuming}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} −\mathrm{3}{x}}={y}\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{3}} =\mathrm{3}{x}+{y}^{\mathrm{3}} \:\left(\mathrm{1}\right) \\ $$$${x}\left({x}−\mathrm{3}\right)\left({x}−\mathrm{9}\right)−\mathrm{8}={x}^{\mathrm{3}} −\mathrm{12}{x}^{\mathrm{2}} +\mathrm{27}{x}−\mathrm{8}= \\ $$$$=\left(\mathrm{3}{x}+{y}^{\mathrm{3}} \right)−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{27}{x}−\mathrm{8}= \\ $$$$=−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{30}{x}+{y}^{\mathrm{3}} −\mathrm{8} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\sqrt[{\mathrm{3}}]{−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{30}{x}+{y}^{\mathrm{3}} −\mathrm{8}}=\mathrm{2}{x}+{y} \\ $$$$−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{30}{x}+{y}^{\mathrm{3}} −\mathrm{8}=\left(\mathrm{2}{x}+{y}\right)^{\mathrm{3}} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \left({y}+\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{4}}{x}\left({y}^{\mathrm{2}} −\mathrm{5}\right)+\mathrm{1}=\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:{x}^{\mathrm{3}} −\mathrm{3}{x}−{y}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \left({y}+\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{4}}{x}\left({y}^{\mathrm{2}} −\mathrm{5}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{match}\:\mathrm{them} \\ $$$$\left({a}\right)\:\frac{\mathrm{3}}{\mathrm{2}}\left({y}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({b}\right)\:\frac{\mathrm{3}}{\mathrm{4}}\left({y}^{\mathrm{2}} −\mathrm{5}\right)=−\mathrm{3} \\ $$$$\left({c}\right)\:−{y}^{\mathrm{3}} =\mathrm{1} \\ $$$$\Rightarrow\:{y}=−\mathrm{1} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{1}\right)\:=\:\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =−\mathrm{2cos}\:\frac{\pi}{\mathrm{9}}\:\approx−\mathrm{1}.\mathrm{87938524} \\ $$$${x}_{\mathrm{2}} =\mathrm{2sin}\:\frac{\pi}{\mathrm{18}}\:\approx.\mathrm{347296355} \\ $$$${x}_{\mathrm{3}} =\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\:\approx\mathrm{1}.\mathrm{53208889} \\ $$$$\mathrm{testing}\:\mathrm{these}\:\mathrm{with}\:\mathrm{the}\:\mathrm{rule}\:\mathrm{for}\:{r}\in\mathbb{R}:\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\Rightarrow\:\mathrm{they}\:\mathrm{all}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$
Commented by Jamshidbek last updated on 13/Aug/21
Thank you! Very good method
$$\mathrm{Thank}\:\mathrm{you}!\:\mathrm{Very}\:\mathrm{good}\:\mathrm{method} \\ $$
Commented by MJS_new last updated on 13/Aug/21
...not sure if it will always work...
$$…\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{it}\:\mathrm{will}\:\mathrm{always}\:\mathrm{work}… \\ $$

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