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Question Number 67136 by ,jamiebots last updated on 23/Aug/19

$$factorize2x^3-1$$

Answered by Cmr 237 last updated on 23/Aug/19

$${2\mathrm{x}}^3-1=2({\mathrm{x}}^3-\frac{1}{2})=\mathrm{p}\left(\mathrm{x}\right)$$$$=2(\mathrm{x}-\sqrt[3]{\frac{1}{2}})({\mathrm{x}}^2+\sqrt[3]{\frac{1}{2}}\mathrm{x}+{\left(\sqrt[3]{\frac{1}{2}}\right)}^2)$$$$\mathrm{or}({\mathrm{x}}^2+\sqrt[3]{\frac{1}{2}}\mathrm{x}+{\left(\sqrt[3]{\frac{1}{2}}\right)}^2)=0\Rightarrow$$$$\mathrm{\Delta }=-3{\left(\sqrt[3]{\frac{1}{2}}\right)}^2={\left(\sqrt{3}\mathrm{i}\sqrt[3]{\frac{1}{2}}\right)}^2$$$${\mathrm{x}}_1=-\frac{\sqrt[3]{\frac{1}{2}}+\sqrt{3}\mathrm{i}\sqrt[3]{\frac{1}{2}}}{2}$$$${\mathrm{x}}_2=-\frac{\sqrt[3]{\frac{1}{2}}-\sqrt{3}\mathrm{i}\sqrt[3]{\frac{1}{2}}}{2}$$$$\mathrm{ainsi},$$$$\mathrm{p}\left(\mathrm{x}\right)=2(\mathrm{x}-{\mathrm{x}}_0)(\mathrm{x}-{\mathrm{x}}_1)(\mathrm{x}-{\mathrm{x}}_2)$$$$\mathbf{avec};{\mathbf{x}}_0=\sqrt[3]{\frac{1}{2}}.$$

Answered by Rasheed.Sindhi last updated on 23/Aug/19

$$2x^3-1={\left({}^3\sqrt{2}x\right)}^3-{\left(1\right)}^3$$$$=({}^3\sqrt{2}x-1)({\left({}^3\sqrt{2}x\right)}^2+\left({}^3\sqrt{2}x\right)\left(1\right)+{\left(1\right)}^2)$$$$=({}^3\sqrt{2}x-1)({}^3\sqrt{4}{x}^2{+}^3\sqrt{2}x+1)$$$$Formula:$$$$a^3-b^3=(a-b)(a^2+ab+b^2)$$$$$$

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