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Question Number 163897 by HongKing last updated on 11/Jan/22

$$\mathrm{if}{\mathrm{x}}^3=1\mathrm{and}\mathrm{x}\ne 1$$$$\mathrm{simplificar}{\left(\frac{\frac{1}{{\mathrm{x}}^4}}{1+{\mathrm{x}}^5}\right)}^3$$

Answered by mr W last updated on 11/Jan/22

$$x^3=1$$$$x^4=x$$$$x^5=x^2=\frac{x^3}{x}=\frac{1}{x}$$$${\left(\frac{\frac{1}{{\mathrm{x}}^4}}{1+{\mathrm{x}}^5}\right)}^3={\left(\frac{1}{x(1+\frac{1}{x})}\right)}^3=\frac{1}{{(1+x)}^3}$$$$=\frac{1}{1+3x+3x^2+x^3}=\frac{1}{3(x^2+x+1)-1}$$$$=\frac{1}{3\times \frac{x^3-1}{x-1}-1}=\frac{1}{3\times 0-1}=-1$$

Commented by HongKing last updated on 11/Jan/22

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{cool}$$

Answered by Rasheed.Sindhi last updated on 14/Jan/22

$$\mathrm{An}\mathrm{other}\mathrm{way}$$$${\mathrm{x}}^3=1\mathrm{and}\mathrm{x}\ne 1;{\left(\frac{\frac{1}{{\mathrm{x}}^4}}{1+{\mathrm{x}}^5}\right)}^3=?$$$${\mathrm{x}}^3-1=0\Rightarrow (\mathrm{x}-1)({\mathrm{x}}^2+\mathrm{x}+1)=0$$$$\Rightarrow {\mathrm{x}}^2+\mathrm{x}+1=0[\because \mathrm{x}\ne 1]$$$$\overline{)\begin{array}{c}{\mathrm{x}}^3=1\wedge {\mathrm{x}}^2+\mathrm{x}+1=0\end{array}}$$$${\mathrm{x}}^3=1:{\left(\frac{\frac{1}{{\mathrm{x}}^4}}{1+{\mathrm{x}}^5}\right)}^3={\left(\frac{\frac{1}{{\mathrm{x}}^3\cdot \mathrm{x}}}{1+{\mathrm{x}}^3\cdot {\mathrm{x}}^2}\right)}^3={\left(\frac{\frac{1}{\mathrm{x}}}{1+{\mathrm{x}}^2}\right)}^3$$$$={\left(\frac{\frac{1}{\mathrm{x}}}{\cancel{1}-\mathrm{x}-\cancel{1}}\right)}^3[\because {\mathrm{x}}^2=-\mathrm{x}-1]$$$$={(-\frac{1}{{\mathrm{x}}^2})}^3=-\frac{1}{{\left({\mathrm{x}}^3\right)}^2}=-1$$

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