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.
 A: I assume we are looking for real solutions. Then our equation is equivalent to the system $x^2+y^2+z^2=1$, $x+y+z=3$. 
Note that because $x^2+y^2+z^2=1$, the absolute values of $x$, $y$, and $z$ are $\le 1$, and they are not all $1$. 
Thus $|x+y+z|\le |x|+|y|+|z| \lt 3$. 
A: (Assuming $x, y, z$ are real.)
How can the sum of two squares be $0$? We know that the square of anything must be positive. Therefore, they both must be zero in order to add up to zero.
$$ x^2 + y^2 + z^2 − 1 = 0 $$
$$ x^2 + y^2 + z^2 = 1 $$
We also have:
$$ x + y + z - 3 = 0 $$
$$ x + y + z = 3 $$
If you want numeric values, you cannot solve the equation (we have 2 equations and 3 variables). However, we can solve for any pair of variables.
If we restrict the range to the reals, there is no solution. You can note this by solving for $x$ and $y$ with respect to $z$, and noting that any solution will be purely imaginary.
A: 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 $x^2+y^2+z^2=1$
