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Question Number 208441 by efronzo1 last updated on 16/Jun/24

Answered by Etimbuk last updated on 16/Jun/24

$${\mathrm{x}}^3-{\mathrm{y}}^3=0$$$${\mathrm{x}}^3={\mathrm{y}}^3\Rightarrow \mathrm{x}=\mathrm{y}$$$$$$$$$$$$$$$$$$$$$$$$\mathrm{the}\mathrm{stated}\mathrm{fraction}\mathrm{will}\mathrm{be}$$$$\frac{{\mathrm{x}}^{2016}+{\left({\mathrm{x}}^2\right)}^{1008}+{\mathrm{y}}^{2016}}{{\left(2\mathrm{x}\right)}^{2016}}=\frac{{2\mathrm{x}}^{2016}+{\mathrm{x}}^{2016}}{2^{2016}{\mathrm{x}}^{2016}}$$$$\frac{{3\mathrm{x}}^{2016}}{2^{2016}{\mathrm{x}}^{2016}}=\frac{3}{2^{2016}}$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by efronzo1 last updated on 16/Jun/24

$$\underset{―}{}$$

Answered by Berbere last updated on 16/Jun/24

$$ify=0\Rightarrow x=0$$$$y\ne 0\Rightarrow {\left(\frac{x}{y}\right)}^2+\left(\frac{x}{y}\right)+1=0z=\frac{x}{y}$$$$\Rightarrow ;z\ne 1z^2+z+1=0\iff z^3=1;z+1=-z^2$$$$\frac{z^{2016}+z^{1008}+1}{{(1+z)}^{2016}}=\frac{(z^{{3)}^{672}}+{\left(z^3\right)}^{336}+1}{{(-z^2)}^{2016}}=\frac{3}{1}=3$$$$$$

Answered by Frix last updated on 16/Jun/24

$$x^2+xy+y^2=0$$$$y=px\wedge x\ne 0\Rightarrow l$$$$p^2+p+1=0$$$$\frac{p^{2016}+p^{1008}+1}{{(p+1)}^{2016}}$$$$p={\mathrm{e}}^{\pm \mathrm{i}\frac{2\pi }{3}}\Rightarrow p+1={\mathrm{e}}^{\pm \mathrm{i}\frac{\pi }{3}}$$$$p^{3n}=1\wedge {(p+1)}^{6n}=1\Rightarrow \frac{p^{2016}+p^{1008}+1}{{(p+1)}^{2016}}=3$$

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