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Question Number 18149 by b.e.h.i.8.3.417@gmail.com last updated on 15/Jul/17

Commented by mrW1 last updated on 16/Jul/17

$$\mathrm{an}\mathrm{other}\mathrm{try}:$$$$\mathrm{let}\mathrm{u}=\mathrm{xyz}$$$$\Rightarrow {\mathrm{x}}^2+\mathrm{xyz}=17\mathrm{x}$$$${\mathrm{x}}^2-17\mathrm{x}+\mathrm{u}=0$$$$\Rightarrow \mathrm{x}=\frac{17\pm \sqrt{{17}^2-4\mathrm{u}}}{2}=\frac{17}{2}\pm \sqrt{{\left(\frac{17}{2}\right)}^2-\mathrm{u}}=\mathrm{a}\pm \sqrt{{\mathrm{a}}^2-\mathrm{u}}$$$$\mathrm{similarly}$$$$\Rightarrow \mathrm{y}=\frac{11\pm \sqrt{{11}^2-4\mathrm{u}}}{2}=\frac{11}{2}\pm \sqrt{{\left(\frac{11}{2}\right)}^2-\mathrm{u}}=\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-\mathrm{u}}$$$$\Rightarrow \mathrm{z}=\frac{13\pm \sqrt{{13}^2-4\mathrm{u}}}{2}=\frac{13}{2}\pm \sqrt{{\left(\frac{13}{2}\right)}^2-\mathrm{u}}=\mathrm{c}\pm \sqrt{{\mathrm{c}}^2-\mathrm{u}}$$$$$$$$\mathrm{u}=\mathrm{xyz}=(\mathrm{a}\pm \sqrt{{\mathrm{a}}^2-\mathrm{u}})(\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-\mathrm{u}})(\mathrm{c}\pm \sqrt{{\mathrm{c}}^2-\mathrm{u}})$$$$......$$

Commented by ajfour last updated on 16/Jul/17

$$\mathrm{Nice}\mathrm{cycle}\mathrm{Sir}.$$

Answered by mrW1 last updated on 16/Jul/17

$$\mathrm{xy}+{\mathrm{x}}^2\mathrm{z}=11\mathrm{x}$$$$({\mathrm{x}}^2-1)\mathrm{z}=11\mathrm{x}-13$$$$\mathrm{z}=\frac{11\mathrm{x}-13}{{\mathrm{x}}^2-1}$$$$\mathrm{y}=11-\frac{(11\mathrm{x}-13)\mathrm{x}}{{\mathrm{x}}^2-1}=\frac{13\mathrm{x}-11}{{\mathrm{x}}^2-1}$$$$$$$$\mathrm{x}+\frac{(11\mathrm{x}-13)(13\mathrm{x}-11)}{{({\mathrm{x}}^2-1)}^2}=17$$$$\mathrm{x}({\mathrm{x}}^4-{2\mathrm{x}}^2+1)+{143\mathrm{x}}^2+143-290\mathrm{x}={17\mathrm{x}}^4-{34\mathrm{x}}^2+17$$$${\mathrm{x}}^5-{2\mathrm{x}}^3+\mathrm{x}+{143\mathrm{x}}^2+143-290\mathrm{x}={17\mathrm{x}}^4-{34\mathrm{x}}^2+17$$$${\mathrm{x}}^5-{17\mathrm{x}}^4-{2\mathrm{x}}^3+{177\mathrm{x}}^2-289\mathrm{x}+126=0$$$$\mathrm{there}\mathrm{are}5\mathrm{roots}:$$$$\mathrm{x}=(-3.64848,0.81019,1.29243,2,16.53586)$$$$\mathrm{e}.\mathrm{g}.\mathrm{with}\mathrm{x}=2$$$$\Rightarrow \mathrm{y}=5$$$$\Rightarrow \mathrm{z}=3$$

Commented by b.e.h.i.8.3.417@gmail.com last updated on 16/Jul/17

$$thanksalotdearmaster.itis$$$$beautifulandsmart.$$

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