x-3-y-3-z-2-x-3-z-3-y-2-y-3-z-3-x-2-obviously-x-y-z-0-1-2-trying-to-totally-solve-it-let-y-px-z-qx-p-3-1-x-3-q-2-x-2-q-3-1-x-3-p-2-x-2-p-3-q-3-x-3-x-2-x-0-y-0-z-0-x-q-2-

FacebookTweetPin

Question Number 176718 by Frix last updated on 25/Sep/22

$$x^3+y^3=z^2$$$$x^3+z^3=y^2$$$$y^3+z^3=x^2$$$$\mathrm{obviously}x=y=z=0\vee \frac{1}{2}$$$$\mathrm{trying}\mathrm{to}\mathrm{totally}\mathrm{solve}\mathrm{it}$$$$\mathrm{let}y=px\wedge z=qx$$$$(p^3+1)x^3=q^2{x}^2$$$$(q^3+1)x^3=p^2{x}^2$$$$(p^3+q^3)x^3=x^2$$$$\Rightarrow x=0\Rightarrow y=0\wedge z=0★$$$$x=\frac{q^2}{p^3+1}$$$$x=\frac{p^2}{q^3+1}$$$$x=\frac{1}{p^3+q^3}$$$$\Rightarrow$$$$\frac{p^2}{q^3+1}=\frac{q^2}{p^3+1}$$$$\frac{p^2}{q^3+1}=\frac{1}{p^3+q^3}$$$$==========$$$$p^5+p^2-q^5-q^2=0$$$$p^5+p^2{q}^3-q^3-1=0$$$$\mathrm{subtracting}\mathrm{both}$$$$p^2(q^3-1)+q^5-q^3+q^2-1=0$$$$(q-1)(p^2(q^2+q+1)+q^4+q^3+q+1)=0$$$$\Rightarrow q=1\Rightarrow p^5+p^2-2=0$$$$(p-1)(p^4+p^3+2p+2)=0$$$$\Rightarrow p=1\Rightarrow x=y=z=\frac{1}{2}★$$$$p^4+p^3+2p+2=0$$$$\mathrm{no}\mathrm{useable}\mathrm{exact}\mathrm{solutions}$$$$p\approx -.975564\pm .528237\mathrm{i}$$$$\Rightarrow x\approx .33635\mp .515329\mathrm{i}$$$$\wedge y\approx -.0559113\pm .680406\mathrm{i}$$$$\wedge z=x★$$$$p\approx .475564\pm 1.18273\mathrm{i}$$$$\Rightarrow x\approx -.586346\pm .562464\mathrm{i}$$$$\wedge y\approx -.944089\mp .426001\mathrm{i}$$$$\wedge z=x★$$$$p^2(q^2+q+1)+q^4+q^3+q+1=0$$$$p^2=-\frac{{(q+1)}^2(q^2-q+1)}{q^2+q+1}$$$$\mathrm{we}\mathrm{can}\mathrm{be}\mathrm{sure}\mathrm{that}(q\ne -1\Rightarrow p=0)$$$$\Rightarrow p^2<0\Rightarrow p=\pm \sqrt{\frac{q^2-q+1}{q^2+q+1}}(q+1)\mathrm{i}$$$$\dots {\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{continue}\mathrm{later}$$

Commented by Tawa11 last updated on 25/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by behi834171 last updated on 26/Sep/22

$$y^3-z^3=z^2-y^2\Rightarrow (y-z)(y^2+z^2+yz)=-(y-z)(y+z)$$$$\Rightarrow x=y=z=0or\frac{1}{2}$$$$y^2+z^2+yz=-(y+z)$$$$z^2+x^2+xz=-(x+z)$$$$x^2+y^2+xy=-(x+y)$$$$\Rightarrow y^2-x^2+z(y-x)=-(y-x)\Rightarrow$$$$\Rightarrow (y-x)(y+x+z+1)=0$$$$\Rightarrow y+x+z=-1$$$$\dots$$$$$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin