Deriving expression for general cubic equation solution.

There isn't a sign mistake in the book. Note that $\omega - \omega^2 = \sqrt{-3}$.

$$\frac{(\omega^2-\omega)}{18}(y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1))$$

$$=\frac{\color{red}{-}\sqrt{-3}}{18}(y_1y_2(y_1-y_2)+y_2y_3(y_2-y_3)+y_1y_3(y_3-y_1))$$

Alternative: \begin{align} &\frac1{54}((y_1+\omega^2y_2+\omega y_3)^3 - (y_1+\omega y_2+\omega^2 y_3)^3)\\ &=\frac1{54}(\omega^2y_2-\omega y_2+\omega y_3 - \omega^2 y_3)((y_1+\omega^2y_2 + \omega y_3)^2+(y_1+\omega^2y_2 + \omega y_3)(y_1+\omega y_2 + \omega^2 y_3)+(y_1+\omega y_2 + \omega^2 y_3)^2)\\ &=\frac{-(\omega - \omega^2)}{54}(y_2-y_3)(3y_1^2+[2(\omega^2y_2 + \omega y_3)+2(\omega y_2 + \omega^2 y_3)+(\omega^2+\omega)(y_2+y_3)]y_1+(\omega^2y_2 + \omega y_3)^2+(\omega^2y_2 + \omega y_3)(\omega y_2 + \omega^2 y_3)+(\omega y_2 + \omega^2 y_3)^2)\\ &= \frac{-\sqrt{-3}}{54}(y_2-y_3)(3y_1^2-3(y_2+y_3)y_1+(\omega y_2^2 + \omega^2 y_3^2+2y_2y_3)+y_2^2+y_3^2+(\omega+\omega^2)y_2y_3+(\omega^2 y_2^2 + \omega y_3^2+2y_2y_3))\\ &=\frac{-\sqrt{-3}}{54}(y_2-y_3)(3y_1^2-3(y_2+y_3)y_1+3y_2y_3)\\ &=\frac{-\sqrt{-3}}{18}(y_2-y_3)(y_1-y_2)(y_1-y_3)\\ &=\frac{\sqrt{-3}}{18}(y_1-y_2)(y_2-y_3)(y_3-y_1)\\ \end{align}

Both $y$ expressions expand to $y_1y_2^2-y_1y_3^2-y_2y_1^2+y_2y_3^2+y_3y_1^2-y_3y_2^2$, and $\omega^2-\omega=2i\Im(\omega)=i\sqrt3$.