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(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Rearranging the terms, we get $y(x^2+x+1) = x^2-x+1$.This simplifies to $(y-1)x^2 + (y+1)x + (y-1) = 0$.For $x$

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$\frac{1}{3}$

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(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Rearranging the terms, we get $y(x^2+x+1) = x^2-x+1$.This simplifies to $(y-1)x^2 + (y+1)x + (y-1) = 0$.For $x$ to be a real number, the discriminant $D$ must be greater than or equal to $0$.$D = (y+1)^2 - 4(y-1)(y-1) \geq 0$.$(y+1)^2 - 4(y-1)^2 \geq 0$.Using the identity $a^2 - b^2 = (a-b)(a+b)$, we get $((y+1) - 2(y-1))((y+1) + 2(y-1)) \geq 0$.$(y+1-2y+2)(y+1+2y-2) \geq 0$.$(3-y)(3y-1) \geq 0$.Multiplying by $-1$ reverses the inequality: $(y-3)(3y-1) \leq 0$.The roots are $y = \frac{1}{3}$ and $y = 3$.Thus, the range of $y$ is $\frac{1}{3} \leq y \leq 3$.The minimum value is $\frac{1}{3}$.

For all real values of $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

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