Question Number 179719 by cortano1 last updated on 01/Nov/22
Commented by Rasheed.Sindhi last updated on 01/Nov/22
$${Is}\:{it}\:\:\sqrt[{\mathrm{5}}]{{x}^{\mathrm{3}} +\mathrm{2}{x}}\:=\sqrt[{\mathrm{3}}]{{x}^{\mathrm{5}} −\mathrm{4}{x}}\:? \\ $$
Commented by mr W last updated on 01/Nov/22
$${x}=\mathrm{0},\:{x}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow{x}=\pm\sqrt{\mathrm{2}} \\ $$
Commented by cortano1 last updated on 01/Nov/22
$$\mathrm{yes}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{0},\:\pm\sqrt{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Nov/22
![((x^3 +2x))^(1/5) =((x^5 −2x))^(1/3) (((x^3 +2x))^(1/5) )^(15) =(((x^5 −2x))^(1/3) )^(15) (x^3 +2x )^3 =(x^5 −2x )^5 x^3 (x^2 +2)^3 =x^5 (x^4 −2)^5 x^3 (x^2 +2)^3 −x^5 (x^4 −2)^5 =0 x^3 {(x^2 +2)^3 −x^2 (x^4 −2)^5 }=0 x=0 ∣ (x^2 +2)^3 −x^2 (x^4 −2)^5 =0 x^3 (x+(2/x))^3 −x^2 ∙x^(10) (x^2 −(2/x^2 ))^5 =0 x^3 (x+(2/x))^3 −x^(12) (x^2 −(2/x^2 ))^5 =0 (x+(2/x))^3 −x^9 (x^2 −(2/x^2 ))^5 =0 [x≠0] Continue...](https://www.tinkutara.com/question/Q179721.png)
$$\sqrt[{\mathrm{5}}]{{x}^{\mathrm{3}} +\mathrm{2}{x}}\:=\sqrt[{\mathrm{3}}]{{x}^{\mathrm{5}} −\mathrm{2}{x}}\: \\ $$$$\left(\sqrt[{\mathrm{5}}]{{x}^{\mathrm{3}} +\mathrm{2}{x}}\:\right)^{\mathrm{15}} =\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{5}} −\mathrm{2}{x}}\:\right)^{\mathrm{15}} \\ $$$$\left({x}^{\mathrm{3}} +\mathrm{2}{x}\:\right)^{\mathrm{3}} =\left({x}^{\mathrm{5}} −\mathrm{2}{x}\:\right)^{\mathrm{5}} \\ $$$${x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} ={x}^{\mathrm{5}} \left({x}^{\mathrm{4}} −\mathrm{2}\right)^{\mathrm{5}} \\ $$$${x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} −{x}^{\mathrm{5}} \left({x}^{\mathrm{4}} −\mathrm{2}\right)^{\mathrm{5}} =\mathrm{0} \\ $$$${x}^{\mathrm{3}} \left\{\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} −{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} −\mathrm{2}\right)^{\mathrm{5}} \right\}=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\mid\:\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} −{x}^{\mathrm{2}} \left({x}^{\mathrm{4}} −\mathrm{2}\right)^{\mathrm{5}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} \left({x}+\frac{\mathrm{2}}{{x}}\right)^{\mathrm{3}} −{x}^{\mathrm{2}} \centerdot{x}^{\mathrm{10}} \left({x}^{\mathrm{2}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{\mathrm{5}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} \left({x}+\frac{\mathrm{2}}{{x}}\right)^{\mathrm{3}} −{x}^{\mathrm{12}} \left({x}^{\mathrm{2}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{\mathrm{5}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({x}+\frac{\mathrm{2}}{{x}}\right)^{\mathrm{3}} −{x}^{\mathrm{9}} \left({x}^{\mathrm{2}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{\mathrm{5}} =\mathrm{0}\:\left[{x}\neq\mathrm{0}\right] \\ $$$$ \\ $$$${Continue}… \\ $$