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Question Number 17985 by tawa tawa last updated on 13/Jul/17

$$\mathrm{solve}\mathrm{simultaneously}.$$$${\mathrm{x}}^3+{\mathrm{y}}^3=35$$$${\mathrm{x}}^4+{\mathrm{y}}^4=97$$

Answered by mrW1 last updated on 13/Jul/17

$$\mathrm{let}\mathrm{u}=\mathrm{x}+\mathrm{y}\mathrm{and}\mathrm{xy}=\mathrm{v}$$$$(\mathrm{x}+\mathrm{y})({\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xy}-3\mathrm{xy})=35$$$$(\mathrm{x}+\mathrm{y})[{(\mathrm{x}+\mathrm{y})}^2-3\mathrm{xy}]=35$$$$\mathrm{u}({\mathrm{u}}^2-3\mathrm{v})=35$$$$\mathrm{v}=\frac{{\mathrm{u}}^2-\frac{35}{\mathrm{u}}}{3}=\frac{{\mathrm{u}}^3-35}{3\mathrm{u}}$$$$$$$${\left({\mathrm{x}}^2\right)}^2+{\left({\mathrm{y}}^2\right)}^2+{2\mathrm{x}}^2{\mathrm{y}}^2-{2\mathrm{x}}^2{\mathrm{y}}^2=97$$$${({\mathrm{x}}^2+{\mathrm{y}}^2)}^2-{2\mathrm{x}}^2{\mathrm{y}}^2=97$$$${({\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xy}-2\mathrm{xy})}^2-{2\mathrm{x}}^2{\mathrm{y}}^2=97$$$${({(\mathrm{x}+\mathrm{y})}^2-2\mathrm{xy})}^2-{2\mathrm{x}}^2{\mathrm{y}}^2=97$$$${({\mathrm{u}}^2-2\mathrm{v})}^2-{2\mathrm{v}}^2=97$$$${({\mathrm{u}}^2-2\frac{{\mathrm{u}}^3-35}{3\mathrm{u}})}^2-2{\left(\frac{{\mathrm{u}}^3-35}{3\mathrm{u}}\right)}^2=97$$$${({\mathrm{u}}^3+70)}^2-2{({\mathrm{u}}^3-35)}^2={873\mathrm{u}}^2$$$${\mathrm{u}}^6+{140\mathrm{u}}^3+4900-{2\mathrm{u}}^6+{140\mathrm{u}}^3-2450={873\mathrm{u}}^2$$$${\mathrm{u}}^6-{280\mathrm{u}}^3+{873\mathrm{u}}^2-2450=0$$$$\Rightarrow \mathrm{u}=-1.391\mathrm{or}5$$$$\Rightarrow \mathrm{v}=9.0322\mathrm{or}6$$$$$$$$\mathrm{x}+\mathrm{y}=\mathrm{u}$$$$\mathrm{xy}=\mathrm{v}$$$${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}+4\mathrm{xy}={\mathrm{u}}^2$$$${(\mathrm{x}-\mathrm{y})}^2={\mathrm{u}}^2-4\mathrm{v}$$$$\mathrm{x}-\mathrm{y}=\pm \sqrt{{\mathrm{u}}^2-4\mathrm{v}}$$$$\Rightarrow \mathrm{x}=\frac{\mathrm{u}\pm \sqrt{{\mathrm{u}}^2-4\mathrm{v}}}{2}$$$$\Rightarrow \mathrm{y}=\frac{\mathrm{u}\mp \sqrt{{\mathrm{u}}^2-4\mathrm{v}}}{2}$$$$$$$$\mathrm{with}\mathrm{u}=-1.391\mathrm{and}\mathrm{v}=9.0322$$$$\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{real}\mathrm{solution}\mathrm{since}$$$${\mathrm{u}}^2-4\mathrm{v}<0$$$$$$$$\mathrm{with}\mathrm{u}=5\mathrm{and}\mathrm{v}=6$$$$\Rightarrow \mathrm{x}+\mathrm{y}=5$$$$\Rightarrow \mathrm{x}-\mathrm{y}=\pm \sqrt{25-24}=\pm 1$$$$\Rightarrow \mathrm{x}=\frac{5\pm 1}{2}=3,2$$$$\Rightarrow \mathrm{y}=\frac{5\mp 1}{2}=2,3$$$$$$$$\Rightarrow (\mathrm{x},\mathrm{y})=(3,2)\mathrm{or}(2,3)$$

Commented by tawa tawa last updated on 13/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}.$$

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