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Question Number 96131 by bemath last updated on 30/May/20

(((x+4)^2 ))^(1/(3 )) + 4 (((x−3)^2 ))^(1/(3 )) + 5 ((x^2 +x−12))^(1/(3 )) = 0

$$\sqrt[{\mathrm{3}\:\:}]{\left({x}+\mathrm{4}\right)^{\mathrm{2}} }\:+\:\mathrm{4}\:\sqrt[{\mathrm{3}\:\:}]{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\:\mathrm{5}\:\sqrt[{\mathrm{3}\:\:}]{{x}^{\mathrm{2}} +{x}−\mathrm{12}}\:=\:\mathrm{0} \\ $$

Answered by john santu last updated on 30/May/20

let ((x+4))^(1/(3 )) = u & ((x−3))^(1/(3 )) = v ⇒u^2 +4v^2 +5uv = 0 (u+4v)(u+v) = 0 { ((u = −v)),((u=−4v)) :} (1) ((x+4))^(1/(3 )) = −((x−3))^(1/(3 )) x+4 = −x+3 ⇒ x=−(1/2) (2) ((x+4))^(1/(3 )) = −4 ((x−3))^(1/(3 )) x+4 = −64x+192 65x = 184 ; x = ((184)/(65)) .

$$\mathrm{let}\:\sqrt[{\mathrm{3}\:\:}]{{x}+\mathrm{4}}\:=\:{u}\:\&\:\sqrt[{\mathrm{3}\:\:}]{{x}−\mathrm{3}}\:=\:{v}\: \\ $$$$\Rightarrow{u}^{\mathrm{2}} +\mathrm{4}{v}^{\mathrm{2}} +\mathrm{5}{uv}\:=\:\mathrm{0} \\ $$$$\left({u}+\mathrm{4}{v}\right)\left({u}+{v}\right)\:=\:\mathrm{0}\: \\ $$$$\begin{cases}{{u}\:=\:−{v}}\\{{u}=−\mathrm{4}{v}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:\sqrt[{\mathrm{3}\:\:}]{{x}+\mathrm{4}}\:=\:−\sqrt[{\mathrm{3}\:\:}]{{x}−\mathrm{3}} \\ $$$${x}+\mathrm{4}\:=\:−{x}+\mathrm{3}\:\Rightarrow\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\sqrt[{\mathrm{3}\:\:}]{{x}+\mathrm{4}}\:=\:−\mathrm{4}\:\sqrt[{\mathrm{3}\:\:}]{{x}−\mathrm{3}} \\ $$$${x}+\mathrm{4}\:=\:−\mathrm{64}{x}+\mathrm{192} \\ $$$$\mathrm{65}{x}\:=\:\mathrm{184}\:;\:{x}\:=\:\frac{\mathrm{184}}{\mathrm{65}}\:. \\ $$

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