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Question Number 197064 by Amidip last updated on 07/Sep/23

Answered by witcher3 last updated on 07/Sep/23

$$\mathrm{x}\sqrt{1+{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^{\frac{2}{3}}}+\mathrm{y}{(1+{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^{\frac{2}{3}})}^{\frac{1}{2}}=\mathrm{a}$$$${\mathrm{a}}^2=({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{x}}^2{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^{\frac{2}{3}}+{\mathrm{y}}^2{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^{\frac{2}{3}})$$$$+2\sqrt{{2\mathrm{x}}^2{\mathrm{y}}^2+{\mathrm{x}}^2{\mathrm{y}}^2({\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^{\frac{2}{3}}+{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^{\frac{2}{3}})}$$$$={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xy}({\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^{\frac{1}{3}}+{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^{\frac{1}{3}})+{\mathrm{x}}^{\frac{4}{3}}{\mathrm{y}}^{\frac{2}{3}}+{\mathrm{y}}^{\frac{4}{3}}{\mathrm{x}}^{\frac{2}{3}}$$$${\mathrm{x}}^2+{\mathrm{y}}^2+{3\mathrm{x}}^{\frac{2}{3}}{\mathrm{y}}^{\frac{4}{3}}+{3\mathrm{y}}^{\frac{2}{3}}{\mathrm{x}}^{\frac{4}{3}}$$$$={({\mathrm{x}}^{\frac{2}{3}}+{\mathrm{y}}^{\frac{2}{3}})}^3={\mathrm{a}}^2\Rightarrow {\mathrm{x}}^{\frac{2}{3}}+{\mathrm{y}}^{\frac{2}{3}}={\mathrm{a}}^{\frac{2}{3}}$$$$$$

Answered by som(math1967) last updated on 07/Sep/23

$$\sqrt{x^{\frac{4}{3}}(x^{\frac{2}{3}}+y^{\frac{2}{3}})}+\sqrt{y^{\frac{4}{3}}(y^{\frac{2}{3}}+x^{\frac{2}{3}})}=a$$$$\Rightarrow x^{\frac{2}{3}}{(x^{\frac{2}{3}}+y^{\frac{2}{3}})}^{\frac{1}{2}}+y^{\frac{2}{3}}{(x^{\frac{2}{3}}+y^{\frac{2}{3}})}^{\frac{1}{2}}=a$$$$\Rightarrow (x^{\frac{2}{3}}+y^{\frac{2}{3}}){(x^{\frac{2}{3}}+y^{\frac{2}{3}})}^{\frac{1}{2}}=a$$$$\Rightarrow {(x^{\frac{2}{3}}+y^{\frac{2}{3}})}^{\frac{3}{2}}=a$$$$\therefore (x^{\frac{2}{3}}+y^{\frac{2}{3}})=a^{\frac{2}{3}}$$

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