x^2-x+1=µ(x^2+x+1) can be written: x^2(1-µ)-x(1+µ)+1-µ=0. We know µ<>1. Using the formula to solve quadratic equations and knowing that the roots of the equation are identical (perfect square), then the square root term must be zero. So (1+µ)^2-4(1-µ)^2=0 (this is b^2-4ac part of the formula), and (1+µ)^2=4(1-µ)^2. There is a perfect square on each side so we can take square roots: 1+µ=+/-2(1-µ). One solution of this is 1+µ=-2+2µ, making µ=3. The other solution is 1+µ=2-2µ, so µ=1/3. [The roots of the equation become x=+/-1.]