Question Number 197064 by Amidip last updated on 07/Sep/23
Answered by witcher3 last updated on 07/Sep/23
$$\mathrm{x}\sqrt{\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }+\mathrm{y}\left(\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{a} \\ $$$$\mathrm{a}^{\mathrm{2}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{y}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \right) \\ $$$$+\mathrm{2}\sqrt{\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \left(\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \right)} \\ $$$$=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2xy}\left(\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right)+\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} \mathrm{y}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{y}^{\frac{\mathrm{4}}{\mathrm{3}}} \mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{3x}^{\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{y}^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{3y}^{\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$=\left(\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\mathrm{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\mathrm{3}} =\mathrm{a}^{\mathrm{2}} \Rightarrow\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{y}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$ \\ $$
Answered by som(math1967) last updated on 07/Sep/23
$$\sqrt{{x}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}\:+\sqrt{{y}^{\frac{\mathrm{4}}{\mathrm{3}}} \left({y}^{\frac{\mathrm{2}}{\mathrm{3}}} +{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}={a} \\ $$$$\Rightarrow{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={a} \\ $$$$\Rightarrow\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={a} \\ $$$$\Rightarrow\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} ={a} \\ $$$$\therefore\left({x}^{\frac{\mathrm{2}}{\mathrm{3}}} +{y}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)={a}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$