the sum of 3 numbers on the second power to be sub 0 This is my exercise 

$x^3 - 3 x^2 + 6 x - a$, the polynom is from $\mathbb{R}[X]$.

And my job is to calculate $x_1^2+x_2^2+x_3^2$. I solved it and I get $-3$ as a result.
My question is: can 3 numbers $a^2+b^2+c^2$ be lower than $0$? Because the power is supposted to make all 3 of them positive.

your answer is correct
your polynomial is : $x^3 - 3 x^2 + 6 x - a  $
and its coefficients are : $a_0=a$, $a_1=6$, $a_2=-3$, $a_3=1$
$\sigma_k = (-1)^k \frac{a_{3-k}}{a3} =(-1)^ka_{3-k} $ for $k=1,2,3$
$$x_1+x_2+x_3=\sigma_1 = -a_2=3$$
$$x_1x_2+x_1x_3+x_2x_3 =\sigma_2 = a_1=6$$
$$(x_1+x_2+x_3)^2= x_1^2+x_2^2+x_3^2+2(x_1x_2+x_1x_3+x_2x_3) \implies x_1^2+x_2^2+x_3^2 = \sigma_1^2 - 2\sigma_2= 9-12 =-3 $$
and to answer your question yes it can be less than $0$ if $a,b,c \in \mathbb{C}$
for example : $a = i, b = 2i, c = -1$   $a^2+b^2+c^2= i^2+4i^2+(-1)^2=-1-4+1=-4$
