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Question Number 67136 by ,jamiebots last updated on 23/Aug/19
factorize 2x^3 −1
$${factorize}\:\mathrm{2}{x}^{\mathrm{3}} −\mathrm{1} \\ $$
Answered by Cmr 237 last updated on 23/Aug/19
2x^3 −1=2(x^3 −(1/2))=p(x) = 2(x−(((1/2) ))^(1/3) )(x^2 +((1/2))^(1/3) x+(((1/2))^(1/3) )^2 ) or (x^2 +((1/2))^(1/3) x+(((1/2))^(1/3) )^2 )=0⇒ Δ=−3(((1/2))^(1/3) )^2 =((√3)i((1/2))^(1/3) )^2 x_1 =−((((1/2))^(1/3) +(√3)i((1/2))^(1/3) )/2) x_2 =−((((1/2))^(1/3) −(√3)i((1/2))^(1/3) )/2) ainsi, p(x)=2(x−x_0 )(x−x_1 )(x−x_2 ) avec;x_0 =((1/2))^(1/3) .
$$\mathrm{2x}^{\mathrm{3}} −\mathrm{1}=\mathrm{2}\left(\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{p}\left(\mathrm{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{2}\left(\mathrm{x}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}\:}\right)\left(\mathrm{x}^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{x}+\left(\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}\right)^{\mathrm{2}} \right) \\ $$$$\:\:\:\mathrm{or}\:\left(\mathrm{x}^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{x}+\left(\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}\right)^{\mathrm{2}} \right)=\mathrm{0}\Rightarrow \\ $$$$\Delta=−\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}\mathrm{i}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} =−\frac{\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}+\sqrt{\mathrm{3}}\mathrm{i}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{2}} =−\frac{\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt{\mathrm{3}}\mathrm{i}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathrm{ainsi}, \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{2}\left(\mathrm{x}−\mathrm{x}_{\mathrm{0}} \right)\left(\mathrm{x}−\mathrm{x}_{\mathrm{1}} \right)\left(\mathrm{x}−\mathrm{x}_{\mathrm{2}} \right) \\ $$$$\boldsymbol{\mathrm{avec}};\boldsymbol{\mathrm{x}}_{\mathrm{0}} =\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}. \\ $$
Answered by Rasheed.Sindhi last updated on 23/Aug/19
2x^3 −1=(^3 (√2) x)^3 −(1)^3 =(^3 (√2) x−1)( (^3 (√2) x)^2 +(^3 (√2) x)(1)+(1)^2 ) =(^3 (√2) x−1)(^3 (√4) x^2 +^3 (√2) x+1 ) Formula: a^3 −b^3 =(a−b)(a^2 +ab+b^2 )
$$\mathrm{2}{x}^{\mathrm{3}} −\mathrm{1}=\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:{x}\right)^{\mathrm{3}} −\left(\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:=\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:{x}−\mathrm{1}\right)\left(\:\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:{x}\right)^{\mathrm{2}} +\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:{x}\right)\left(\mathrm{1}\right)+\left(\mathrm{1}\right)^{\mathrm{2}} \:\right) \\ $$$$\:\:\:\:=\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:{x}−\mathrm{1}\right)\left(^{\mathrm{3}} \sqrt{\mathrm{4}}\:{x}^{\mathrm{2}} +^{\mathrm{3}} \sqrt{\mathrm{2}}\:{x}+\mathrm{1}\:\right) \\ $$$$\:\:{Formula}: \\ $$$$\:\:\:\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right) \\ $$$$ \\ $$

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