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Question Number 163163 by mathlove last updated on 04/Jan/22

x^9 −2022x^3 +(√(2021))=0  x={?}

$${x}^{\mathrm{9}} −\mathrm{2022}{x}^{\mathrm{3}} +\sqrt{\mathrm{2021}}=\mathrm{0} \\ $$$${x}=\left\{?\right\} \\ $$

Commented by mr W last updated on 04/Jan/22

what do you want to say with {?}?  do you just mean x=? ?

$${what}\:{do}\:{you}\:{want}\:{to}\:{say}\:{with}\:\left\{?\right\}? \\ $$$${do}\:{you}\:{just}\:{mean}\:{x}=?\:? \\ $$

Answered by mr W last updated on 04/Jan/22

let x^3 =t  t^3 −2022t+(√(2021))=0  t=2(√(674)) sin (((2kπ)/3)+(1/3)sin^(−1) (1/( 1348))(√((2021)/(674)))) (k=0,1,2)  ⇒x=((2(√(674)) sin (((2kπ)/3)+(1/3)sin^(−1) (1/( 1348))(√((2021)/(674))))))^(1/3)

$${let}\:{x}^{\mathrm{3}} ={t} \\ $$$${t}^{\mathrm{3}} −\mathrm{2022}{t}+\sqrt{\mathrm{2021}}=\mathrm{0} \\ $$$${t}=\mathrm{2}\sqrt{\mathrm{674}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\mathrm{1348}}\sqrt{\frac{\mathrm{2021}}{\mathrm{674}}}\right)\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}\sqrt{\mathrm{674}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\mathrm{1348}}\sqrt{\frac{\mathrm{2021}}{\mathrm{674}}}\right)} \\ $$

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