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x-1-3-x-1-3-2-2-x-1-3-x-1-3-2-2-x-3-4-




Question Number 75753 by Emmanuel_N last updated on 16/Dec/19
((((x^(1/3) +x^(−1/3) )^2 −2)/((x^(1/3) +x^(−1/3) )^2 +2))−x)^(3/4)
$$\left(\frac{\left(\mathrm{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{x}^{−\mathrm{1}/\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}}{\left(\mathrm{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{x}^{−\mathrm{1}/\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{2}}−\mathrm{x}\right)^{\mathrm{3}/\mathrm{4}} \\ $$
Answered by $@ty@m123 last updated on 16/Dec/19
Let y=x^(1/3)   =[(((y+(1/y))^2 −2)/((y−(1/y))^2 +2)) −x]^(3/4)   =[((y^2 +(1/y^2 ))/(y^2 +(1/y^2 ))) −x]^(3/4)   =(1−x)^(3/4)
$${Let}\:{y}={x}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\left[\frac{\left({y}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}}{\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} +\mathrm{2}}\:−{x}\right]^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=\left[\frac{{y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }}{{y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }}\:−{x}\right]^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 17/Dec/19
but the original question is  =[(((y+(1/y))^2 −2)/((y+(1/y))^2 +2)) −x]^(3/4)
$${but}\:{the}\:{original}\:{question}\:{is} \\ $$$$=\left[\frac{\left({y}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}}{\left({y}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} +\mathrm{2}}\:−{x}\right]^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$

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