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Question Number 168131 by daus last updated on 04/Apr/22

Commented by daus last updated on 04/Apr/22

help me to solve above equation

$${help}\:{me}\:{to}\:{solve}\:{above}\:{equation} \\ $$

Answered by benhamimed last updated on 04/Apr/22

(d)⇔ { ((x^3 =t)),((t^2 −2t−8=0)) :} t=−2 ;t=4 S={−(2)^(1/3) ;(4)^(1/3) } (e)→x^6 −6x^3 −16=0 ⇔ { ((x^3 =t)),((t^2 −6t−16=0)) :} t=−2;8 S={−(2)^(1/3) ;2}

$$\left({d}\right)\Leftrightarrow\begin{cases}{{x}^{\mathrm{3}} ={t}}\\{{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{8}=\mathrm{0}}\end{cases} \\ $$$${t}=−\mathrm{2}\:;{t}=\mathrm{4} \\ $$$${S}=\left\{−\sqrt[{\mathrm{3}}]{\mathrm{2}};\sqrt[{\mathrm{3}}]{\mathrm{4}}\right\} \\ $$$$\left({e}\right)\rightarrow{x}^{\mathrm{6}} −\mathrm{6}{x}^{\mathrm{3}} −\mathrm{16}=\mathrm{0} \\ $$$$\Leftrightarrow\begin{cases}{{x}^{\mathrm{3}} ={t}}\\{{t}^{\mathrm{2}} −\mathrm{6}{t}−\mathrm{16}=\mathrm{0}}\end{cases} \\ $$$${t}=−\mathrm{2};\mathrm{8} \\ $$$${S}=\left\{−\sqrt[{\mathrm{3}}]{\mathrm{2}};\mathrm{2}\right\} \\ $$

Answered by MJS_new last updated on 04/Apr/22

x^6 −2x^3 −8=0 (x^3 −4)(x^3 +2)=0 x_1 =(4)^(1/3) x_(2, 3) =−((1±(√3)i)/( (2)^(1/3) )) x_4 =−(2)^(1/3) x_(5, 6) =((1±(√3)i)/( (4)^(1/3) )) x^2 =((2(3x^3 +8))/x^4 ) x^6 −6x^3 −16=0 (x−2)(x^2 +2x+4)(x^3 +2)=0 x_1 =2 x_(2, 3) =−1±(√3)i x_4 =−(2)^(1/3) x_(5, 6) =((1±(√3)i)/( (4)^(1/3) ))

$${x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{8}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{4}\right)\left({x}^{\mathrm{3}} +\mathrm{2}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =−\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}\mathrm{i}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}} \\ $$$${x}_{\mathrm{4}} =−\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$${x}_{\mathrm{5},\:\mathrm{6}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}\mathrm{i}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}} \\ $$$$ \\ $$$$ \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{2}\left(\mathrm{3}{x}^{\mathrm{3}} +\mathrm{8}\right)}{{x}^{\mathrm{4}} } \\ $$$${x}^{\mathrm{6}} −\mathrm{6}{x}^{\mathrm{3}} −\mathrm{16}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)\left({x}^{\mathrm{3}} +\mathrm{2}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{2} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =−\mathrm{1}\pm\sqrt{\mathrm{3}}\mathrm{i} \\ $$$${x}_{\mathrm{4}} =−\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$${x}_{\mathrm{5},\:\mathrm{6}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}\mathrm{i}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}} \\ $$

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