The coefficient of x³ in the expansion of ${(1-\mathrm{x}+{\mathrm{x}}^2)}^6$ is

$\begin{array}{rl} & {(1-\mathrm{x}+{\mathrm{x}}^2)}^6=[1-\mathrm{x}(1-\mathrm{x}){]}^6\\ & {=}^6{\mathrm{C}}_0{-}^6{\mathrm{C}}_1\mathrm{x}(1-\mathrm{x}){+}^6{\mathrm{C}}_2{\mathrm{x}}^2(1-\mathrm{x}{)}^2{-}^6{\mathrm{C}}_3{\mathrm{x}}^3(1-\mathrm{x}{)}^3+\dots \text{to }7 \text{terms}\\ & {=}^6{\mathrm{C}}_0{-}^6{\mathrm{C}}_1\mathrm{x}(1-\mathrm{x}){+}^6{\mathrm{C}}_2{\mathrm{x}}^2(1-2\mathrm{x}+{\mathrm{x}}^2){-}^6{\mathrm{C}}_3{\mathrm{x}}^3(1-3\mathrm{x}+3{\mathrm{x}}^2-{\mathrm{x}}^3)+\dots \text{to }7 \text{terms}\end{array}$

$\therefore$Coefficient of ${\mathrm{x}}^3=-2{\cdot }^6{\mathrm{C}}_2{-}^6{\mathrm{C}}_3$ (collecting coefficients of x3 from each term)

$=-2\frac{6!}{2!4!}-\frac{6!}{3!3!}=-50$