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Question Number 165483 by mathlove last updated on 02/Feb/22

Commented by MJS_new last updated on 04/Feb/22

$$\mathrm{for}x=y=z\mathrm{there}\mathrm{are}3\mathrm{solutions}\mathrm{in}\mathbb{C}$$$$\mathrm{for}x=y\ne z\mathrm{there}\mathrm{are}9\mathrm{solutions}\mathrm{in}\mathbb{C}$$$$(\mathrm{and}\mathrm{of}\mathrm{course}x,y,z\mathrm{are}\mathrm{interchangeable})$$$$\mathrm{for}x\ne y\ne z\mathrm{I}\mathrm{had}\mathrm{no}\mathrm{time}\mathrm{yet}$$

Answered by alephzero last updated on 02/Feb/22

$$x=y=z$$

Commented by mathlove last updated on 02/Feb/22

$$how?$$

Commented by alephzero last updated on 02/Feb/22

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{obvious}$$

Answered by Rasheed.Sindhi last updated on 02/Feb/22

$$\{\begin{array}{l}12x^3+12y^3+9z=2......\left(i\right)\\ 12y^3+12z^3+9x=2......\left(ii\right)\\ 12z^3+12x^3+9y=2...\left(iii\right)\end{array}$$$$\left(i\right)-\left(iii\right):12y^3-12z^3+9z-9y=0$$$$12(y-z)(y^2+yz+z^2)-9(y-z)=0$$$$(y-z)(12(y^2+yz+z^2)-9)=0$$$$y=z\mid y^2+yz+z^2=\frac{3}{4}$$$$Similarly,$$$$\left(iii\right)-\left(ii\right)\Rightarrow z=x$$$$\therefore x=y=z$$

Commented by mr W last updated on 03/Feb/22

$$x=y=zisonlyonepossibititywhich$$$$givesus3solutions.$$$$whataboutotherpossibilities:$$$$x=y\ne z$$$$x=z\ne y$$$$y=z\ne x$$$$x\ne y\ne z$$$$ithinkwecouldgetmoresolutions$$$$fromthesepossibilities.$$

Commented by Rasheed.Sindhi last updated on 03/Feb/22

$$\{\begin{array}{l}y-z=0\mid y^2+yz+z^2=\frac{3}{4}....\left(i\right)\\ x-y=0\mid x^2+xy+y^2=\frac{3}{4}...\left(ii\right)\\ z-x=0\mid z^2+zx+x^2=\frac{3}{4}...\left(iii\right)\end{array}$$$$\left(ii\right)-\left(i\right):x^2+xy-yz-z^2=0$$$$(x-z)(x+z)+y(x-z)=0$$$$(x-z)(x+y+z)=0$$$$x=z\mid x+y+z=0$$

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