I challenge you to solve this properly!

There is a problem with this problem. Since $a_1$ is a positive real, $b_1$ can be taken as large as one pleases (with the appropriate choice of $a_1$ , which also increases), without changing the value of ${a_1}^2-2{b_1}^2$ . That means the value of $[x+a_1]+[y+b_1]+[z]$ does not necessarily remain constant for a given choice of $x,y,z$ , and can take infinitely many values. This is problematic. I just simply chose their values alongside $x,y,z$ , and my answer happened to match the given answer. To start, I choose $c=-\sqrt[3]{|p|}$ , and $a$ as the primitive root. Since $a,b,c$ all have an equal modulus of $\sqrt[3]{|p|}$ , this modulus can be cancelled from the numerator and denominator of the expression $\Big|\frac{xa+yb+zc}{xb+yc+za}\Big|$ . Therefore, I can choose to write it as $\Big|\frac{x cis(\pi/3)+y cis(-\pi/3)-z}{x cis(-\pi/3)-y+z cis(\pi/3)}\Big|$ . Next, I take $x=1, y=cis(-\pi/3), z=cis(\pi/3)$ . If you check my work, you might find that the expression resolves to $\Big|\frac{-cis(\pi/3)}{cis(-\pi/3)}\Big|=1$ . Next, since $\omega$ and $\omega^2$ are conjugates, their coefficients must be equal since the right side of the equation is zero (look at the imaginary part). Hence, ${a_1}^2-2{b_1}^2=[x]+[y]+[z]=1+[cis(-\pi/3)]+[cis(\pi/3)]=1$ . So I choose $a_1=3$ and $b_1=2$ . Thus, the equation is satisfied, as $1+\omega +\omega^2=0$ . And so, $[x+a_1]+[y+b_1]+[z]=[1+3]+[cis(-\pi/3)+2]+[cis(\pi/3)]=6$ .

The term inside the modulus turns out to be a complex root of unity hence its modulus becomes 1.Now the eqn reduces to 1+(a1^2-2b1^2)w+([x]+[y]+[z])w^2 = 0. Comparing this equation with 1+w+w^2=0 ,we get a1^2-2b1^2=1,[x]+[y]+[z]=1. from first equation since b1 is prime the only solution left for a1 & b1 is 3 & 2 respectively. Therefore the required expression [x+a1] +[y+b1]+[z] = [x]+[y]+[z] +a1+b1(because a1 and b1 are integers) which is equal to 1+3+2=6