A Too Easy One in Fact!

With $x^2+x+1 = 0$ , we multiply $x-1$ for both sides, and gives $x^3-1 = 0$ . Now, since $\alpha^{62} = (1^{20})\alpha^2$ , and $\beta^{62} = (1^{20})\beta^2$ , we resolve for $\alpha, \beta$ , which gives $\alpha = \frac{-1+i\sqrt{3}}{2};\:\beta = \frac{-1-i\sqrt{3}}{2}$ and $\beta^2 = \alpha;\: \alpha^2 = \beta$ , hence we will come to the same equation, the $x^2+x+1 = 0$ , where the $gcd(1,1,1) = 1$ , but since we require the equation with $gcd(a,b,c) = 22$ , we just multiply 22 into the original equation, which gives $22x^2+22x+22 = 0$ , hence, $a+b+c = 22+22+22 = 66$