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Question Number 75814 by Master last updated on 18/Dec/19

Commented by Master last updated on 18/Dec/19

xyz=?

$$\mathrm{xyz}=? \\ $$

Answered by mr W last updated on 18/Dec/19

(i)+(ii):  2x^2 +2y−2x=0  ⇒y=x−x^2   ⇒z=1+2x−y=1+x+x^2   put this into (iii):  x^4 +(x+x^2 )(x−x^2 )=1  x^2 =1  ⇒x=1 or −1  ⇒y=0 or −2  ⇒z=3 or 1    ⇒xyz=0 or 2

$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{2}{x}=\mathrm{0} \\ $$$$\Rightarrow{y}={x}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow{z}=\mathrm{1}+\mathrm{2}{x}−{y}=\mathrm{1}+{x}+{x}^{\mathrm{2}} \\ $$$${put}\:{this}\:{into}\:\left({iii}\right): \\ $$$${x}^{\mathrm{4}} +\left({x}+{x}^{\mathrm{2}} \right)\left({x}−{x}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{1}\:{or}\:−\mathrm{1} \\ $$$$\Rightarrow{y}=\mathrm{0}\:{or}\:−\mathrm{2} \\ $$$$\Rightarrow{z}=\mathrm{3}\:{or}\:\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow{xyz}=\mathrm{0}\:{or}\:\mathrm{2} \\ $$

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