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x-3-x-2-x-1-3-Find-the-value-of-x-Help-me-please-

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Question Number 215177 by Ismoiljon_008 last updated on 30/Dec/24
x^3 + x^2 + x = − (1/3) Find the value of x ! Help me, please
$$ \\ $$$$\:\:\:{x}^{\mathrm{3}} \:+\:{x}^{\mathrm{2}} \:+\:{x}\:=\:−\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\mathcal{F}{ind}\:{the}\:{value}\:{of}\:\:{x}\:\:!\: \\ $$$$ \\ $$$$\:\:\:\mathcal{H}{elp}\:{me},\:\:{please} \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 30/Dec/24
3x^3 +3x^2 +3x +1=0 2x^3 + (x+1)^3 =0 ((2)^(1/3) x + x+1 )( {(4)^(1/3) x^2 −x^2 (2)^(1/3) −(2)^(1/3) x + x^2 +2x+1}>0)=0 −x((2)^(1/3) +1 )=1 ⇒ x= ((−1)/(1+ (2)^(1/3) ))
$$\:\:\:\:\:\:\:\:\mathrm{3}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}\:+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{x}^{\mathrm{3}} \:+\:\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:{x}\:+\:{x}+\mathrm{1}\:\right)\left(\:\left\{\sqrt[{\mathrm{3}}]{\mathrm{4}}\:{x}^{\mathrm{2}} −{x}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\mathrm{2}}\:{x}\:+\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right\}>\mathrm{0}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:−{x}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\:+\mathrm{1}\:\right)=\mathrm{1}\:\Rightarrow\:{x}=\:\frac{−\mathrm{1}}{\mathrm{1}+\:\sqrt[{\mathrm{3}}]{\mathrm{2}}} \\ $$
Answered by Ghisom last updated on 31/Dec/24
x^3 +x^2 +x+(1/3)=0 x=t−(1/3) t^3 +(2/3)t+(2/(27))=0 Cardano′s Formula gives t_1 =−(2^(2/3) /3)+(2^(1/3) /3) ⇒ x_1 =−((2^(2/3) −2^(1/3) +1)/3) t_2 =−(2^(2/3) /3)ω+(2^(1/3) /3)ω^2 ⇒ x_2 =((2^(2/3) −2^(1/3) −2)/6)+(((2^(2/3) +2^(1/3) )(√3))/6)i t_3 =−(2^(2/3) /3)ω^2 +(2^(1/3) /3)ω ⇒ x_3 =((2^(2/3) −2^(1/3) −2)/6)−(((2^(2/3) +2^(1/3) )(√3))/6)i
$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${x}={t}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{3}}{t}+\frac{\mathrm{2}}{\mathrm{27}}=\mathrm{0} \\ $$$$\mathrm{Cardano}'\mathrm{s}\:\mathrm{Formula}\:\mathrm{gives} \\ $$$${t}_{\mathrm{1}} =−\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}_{\mathrm{1}} =−\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} −\mathrm{2}^{\mathrm{1}/\mathrm{3}} +\mathrm{1}}{\mathrm{3}} \\ $$$${t}_{\mathrm{2}} =−\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} }{\mathrm{3}}\omega+\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}}\omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}_{\mathrm{2}} =\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} −\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\mathrm{2}}{\mathrm{6}}+\frac{\left(\mathrm{2}^{\mathrm{2}/\mathrm{3}} +\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right)\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{i} \\ $$$${t}_{\mathrm{3}} =−\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} }{\mathrm{3}}\omega^{\mathrm{2}} +\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}}\omega \\ $$$$\:\:\:\:\:\Rightarrow\:{x}_{\mathrm{3}} =\frac{\mathrm{2}^{\mathrm{2}/\mathrm{3}} −\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\mathrm{2}}{\mathrm{6}}−\frac{\left(\mathrm{2}^{\mathrm{2}/\mathrm{3}} +\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right)\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{i} \\ $$

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