8. The real numbers $x, y$ and $z$ are a solution $(x, y, z)$ of the equation $\left(x^{2}-9\right)^{2}+\left(y^{2}-4\right)^{2}+\left(z^{2}-1\right)^{2}=0$. How many different possible values are there for $x+y+z ?$
8. 7 Since squares of real numbers are non-negative, the sum can only be 0 if each expression in brackets is zero. Therefore the solutions of the equation are $x= \pm 3, y= \pm 2$ and $z= \pm 1$. We observe that the maximum and minimum values for $x+y+z$ are 6 and -6 , and that since $x+y+z$ is the sum of one even and two odd numbers, that $x+y+z$ itself will be even.
It suffices to show that each even total between +6 and -6 can be attained.
$$
\begin{array}{ll}
(+3)+(+2)+(+1)=+6 & (+3)+(+2)+(-1)=+4 \\
(+3)+(-2)+(+1)=+2 & (+3)+(-2)+(-1)=0 \\
(-3)+(+2)+(-1)=-2 & (-3)+(-2)+(+1)=-4 \\
(-3)+(-2)+(-1)=-6 &
\end{array}
$$

Hence there are seven possible values for $x+y+z$.