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I saw it on AoPS, very great: https://artofproblemsolving.com/community/c6h1274759p6726915 JunBo-Yang used his own substitution for the condition: $a^{\,2}+ b^{\,2}+ c^{\,2}+ 3\,abc= 6$ , then: $$a= \frac{\left ( 3- \sqrt{3} \right )x+ \left ( 3+ \sqrt{3} \right )y}{\sqrt{\left ( 4\,y+ z+ x \right )\left ( 4\,x+ y+ z \right )}}$$ $$b= \frac{\left ( 3- \sqrt{3} \right )z+ \left ( 3+ \sqrt{3} \right )x}{\sqrt{\left ( 4\,x+ y+ z \right )\left ( 4\,z+ x+ y \right )}}$$ $$c= \frac{\left ( 3- \sqrt{3} \right )y+ \left ( 3+ \sqrt{3} \right )z}{\sqrt{\left ( 4\,z+ x+ y \right )\left ( 4\,y+ z+ x \right )}}$$ By the same way, the condition: $a^{\,2}+ b^{\,2}+ c^{\,2}+ k\,abc= k+ 3$ can be substituted by: $$a= \frac{\left ( k- \sqrt{k} \right )x+ \left ( k+ \sqrt{k} \right )y}{\sqrt{\left ( \left ( k+ 1 \right )\,y+ z+ x \right )\left ( \left ( k+ 1 \right )\,x+ y+ z \right )}}$$ $$b= \frac{\left ( k- \sqrt{k} \right )z+ \left ( k+ \sqrt{k} \right )x}{\sqrt{\left ( \left ( k+ 1 \right )\,x+ y+ z \right )\left ( \left ( k+ 1 \right )\,z+ x+ y \right )}}$$ $$c= \frac{\left ( k- \sqrt{k} \right )y+ \left ( k+ \sqrt{k} \right )z}{\sqrt{\left ( \left ( k+ 1 \right )\,z+ x+ y \right )\left ( \left ( k+ 1 \right )\,y+ z+ x \right )}}$$ But I can't know how to make it like that, I tried to solve the system of a cyclic sum and a cyclic product like: $$\left\{\begin{matrix} a^{\,2}+ b^{\,2}+ c^{\,2} & = & X & \left ( 1 \right )\\ 3\,abc & = & 6- X & \left ( 2 \right ) \end{matrix}\right.$$ We can wirte $\left ( 1 \right )$ and $\left ( 2 \right )$ homogeneous, easily. But I can't combine both of them, it's pretty hard. If you can, give me some help. I am continuing... Thanks!

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If your substitution is true it should be true for $k=0$, but it's not so.

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