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Question Number 152772 by Rankut last updated on 01/Sep/21

if  ((((x−2))^(1/3) +2))^(1/3)  + ((2−((x−2))^(1/3) ))^(1/3) =2   then  find the value of   (√(198x^4 −868x^3 −229x^2 +200x))

$$\boldsymbol{{if}}\:\:\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}−\mathrm{2}}+\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}−\mathrm{2}}}=\mathrm{2} \\ $$$$\:\boldsymbol{\mathrm{then}}\:\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\sqrt{\mathrm{198}\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\mathrm{868}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{229}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{200}\boldsymbol{\mathrm{x}}} \\ $$

Commented by Rankut last updated on 01/Sep/21

any help please

$${any}\:{help}\:{please} \\ $$

Commented by MJS_new last updated on 01/Sep/21

typo?  there′s a similae question... just scroll down  to find it

$$\mathrm{typo}? \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{similae}\:\mathrm{question}...\:\mathrm{just}\:\mathrm{scroll}\:\mathrm{down} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{it} \\ $$

Answered by mr W last updated on 01/Sep/21

let t=((x−2))^(1/3)   ((2+t))^(1/3) +((2−t))^(1/3) =2  2+t+2−t+3×2((4−t^2 ))^(1/3) =8  3((4−t^2 ))^(1/3) =2  27(4−t^2 )=8  t=±(√(4−(8/(27))))=±((10(√3))/9)  x=2+t^3 =2±((1000(√3))/(243))  ......

$${let}\:{t}=\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}+{t}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−{t}}=\mathrm{2} \\ $$$$\mathrm{2}+{t}+\mathrm{2}−{t}+\mathrm{3}×\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{4}−{t}^{\mathrm{2}} }=\mathrm{8} \\ $$$$\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{4}−{t}^{\mathrm{2}} }=\mathrm{2} \\ $$$$\mathrm{27}\left(\mathrm{4}−{t}^{\mathrm{2}} \right)=\mathrm{8} \\ $$$${t}=\pm\sqrt{\mathrm{4}−\frac{\mathrm{8}}{\mathrm{27}}}=\pm\frac{\mathrm{10}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$${x}=\mathrm{2}+{t}^{\mathrm{3}} =\mathrm{2}\pm\frac{\mathrm{1000}\sqrt{\mathrm{3}}}{\mathrm{243}} \\ $$$$...... \\ $$

Commented by Rankut last updated on 01/Sep/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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