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Question Number 192914 by ajfour last updated on 30/May/23

$$x^2(x^2-1)={(1-\frac{c}{x})}^3+{\left(\frac{c}{x}\right)}^3$$

Answered by a.lgnaoui last updated on 31/May/23

$${\mathrm{x}}^2({\mathrm{x}}^2-1)=\frac{{(\mathrm{x}-\mathrm{c})}^3+{\mathrm{c}}^3}{{\mathrm{x}}^3}$$$${\mathrm{x}}^5({\mathrm{x}}^2-1)={\mathrm{x}}^3-3\mathrm{cx}(\mathrm{x}-\mathrm{c})$$$${\mathrm{x}}^4({\mathrm{x}}^2-1)={\mathrm{x}}^2-3\mathrm{c}(\mathrm{x}-\mathrm{c})$$$${\mathrm{x}}^6-{\mathrm{x}}^4={(\mathrm{x}-\frac{3\mathrm{c}}{2})}^2+\frac{{3\mathrm{c}}^2}{4}$$$${[(\mathrm{x}-\frac{3\mathrm{c}}{2})+\frac{3\mathrm{c}}{2}]}^6={[(\mathrm{x}-\frac{3\mathrm{c}}{2})+\frac{3\mathrm{c}}{2}]}^4+{(\mathrm{x}-\frac{3\mathrm{c}}{2})}^2+\frac{{3\mathrm{c}}^2}{4}$$$$\mathbf{X}=\mathbf{x}-\mathbf{a}\mathbf{avec}\frac{3\mathbf{c}}{2}=\mathbf{a}$$$${(\mathbf{X}+\mathbf{a})}^6={(\mathbf{X}+\mathbf{a})}^4+{\mathbf{X}}^2+\frac{\mathbf{ac}}{2}$$$${(\mathbf{X}+\mathbf{a})}^6-{(\mathbf{X}+\mathbf{a})}^4-{[(\mathbf{X}+\mathbf{a})-\mathbf{a}]}^2+\frac{\mathbf{ac}}{2}$$$$\mathbf{X}+\mathbf{a}=\mathbf{Z}$$$$$$$${\mathbf{Z}}^6-{\mathbf{Z}}^4-({\mathbf{Z}}^2-2\mathbf{aZ}+{\mathbf{a}}^2)-\frac{\mathbf{ac}}{2}=0$$$${\mathbf{Z}}^6-{\mathbf{Z}}^4-{\mathbf{Z}}^2+2\mathbf{aZ}-\frac{\mathbf{a}(2\mathbf{a}+\mathbf{c})}{2}=0$$$$$$$$\mathbf{soit}{\mathbf{Z}}^6-{\mathbf{Z}}^4-{\mathbf{Z}}^2+3\mathbf{cZ}-3{\mathbf{c}}^2=0$$$$$$$$\mathbf{Z}=\mathbf{X}+\mathbf{a}=\mathbf{x}$$$$$$$$$$$${\mathbf{x}}^6-{\mathbf{x}}^4-{\mathbf{x}}^2+3\mathbf{c}(\mathbf{x}-\mathbf{c})=0$$$$...........$$

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