<s>[INST] Finding all solutions of the equation $(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$. Find all solutions of the equation
$(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$.    

not getting any clue.can somebody help me please.thanks.
 [/INST] If $x,y,z$ are real, we can set $x=\cos A\cos B,y=\cos A\sin B,z=\sin A$
Now, $$\cos A\cos B+\cos A\sin B+\sin A=\cos A\sqrt2\cos(B-\frac\pi4)+\sin A\le \sqrt2\cos A+\sin A$$
Let $\sqrt2=R\cos C,1=R\sin C$ where $R>0\implies R^2=2+1=3,R=\sqrt3, \cos C=\sqrt\frac23$
$\implies \sqrt2\cos A+\sin A=\sqrt3\cos(A-\arccos\sqrt\frac23)\le \sqrt 3$
So, $x+y+z\le \sqrt3$ which occurs when $A-\arccos\sqrt\frac23=2n\pi\implies \cos A=\sqrt\frac23$ and $B=\frac \pi4$
So there is no real solution for $x+y+z=3$ if $x^2+y^2+z^2=1$
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