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Question Number 215282 by arkanmathematics63 last updated on 02/Jan/25

$$Showthatx=64andy=729$$$$\sqrt[3]{x}+\sqrt[3]{y}=13$$$$\sqrt{x}+\sqrt{y}=35$$

Commented by Ghisom last updated on 02/Jan/25

$$\mathrm{to}‘‘\mathrm{show}\mathrm{that}x=64\mathrm{and}y={729}^{″}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{enough}$$$$\mathrm{to}\mathrm{insert}:$$$$\sqrt[3]{64}+\sqrt[3]{729}=4+9=13$$$$\sqrt{64}+\sqrt{729}=8+27=35$$

Commented by TonyCWX08 last updated on 02/Jan/25

$$Yougotapoint...................................$$

Commented by Ghisom last updated on 02/Jan/25

$$...\mathrm{btw}\mathrm{there}\mathrm{are}\mathrm{no}\mathrm{solutions}\notin \mathbb{R}$$

Answered by TonyCWX08 last updated on 02/Jan/25

$$Letu=x^6,v=y^6$$$$u^2+v^2=13\Rightarrow p_2=13$$$$u^3+v^3=35\Rightarrow p_3=35$$$$$$$$By{Newton}^{\prime }sIdentity,$$$$p_1=e_1$$$$$$$$p_2=e_1^2-2e_2\Rightarrow e_2=\frac{p_1^2-13}{2}$$$$$$$$p_3=e_1^3-2e_1{e}_2+3e_3$$$$35=p_1^3-2\left(p_1\right)\left(\frac{p_1^2-13}{2}\right)$$$$\frac{39p_1-p_1^3}{2}=35$$$$39p_1-p_1^3=70$$$$p_1=-7,2,5$$$$$$$$u+v=-7$$$$u+v=2$$$$u+v=5$$$$$$$$Case1:$$$$u+v=-7\Rightarrow u=-7-v$$$${(-7-v)}^2+v^2=13\Rightarrow NoSolutioninrealworld$$$$$$$$Case2:$$$$u+v=2\Rightarrow u=2-v$$$${(2-v)}^2+v^2=13$$$$\Rightarrow (u,v)=(2,3)or(3,2)$$$$\Rightarrow (x,y)=(64,729)or(729,64)$$$$$$$$Case3:$$$$u+v=5\Rightarrow u=5-v$$$${(5-v)}^2+v^2=13$$$$\Rightarrow (u,v)=(\frac{2-\sqrt{22}}{2},\frac{2+\sqrt{22}}{2})or(\frac{2+\sqrt{22}}{2},\frac{2-\sqrt{22}}{2})$$$$\Rightarrow (x,y)=Nosolutioninrealworld$$

Commented by arkanmathematics63 last updated on 12/Jan/25

$$WOW!$$$$NiceSolution.ThankyouSir.$$

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