The coefficient of $x^4$ in the expansion of ${(1-x-x^2+x^3)}^6$ is 

$\begin{array}{l}{\left(1-x-x^2-x^3\right)}^6={\left((1-x)-x^2(1-x)\right)}^6\\ ={\left((1-x)\left(1-x^2\right)\right)}^6\\ =(1-x{)}^6{\left(1-x^2\right)}^6\\ =\left({ }^6{C}_0{-}^6{C}_{1}x{+}^6{C}_2{x}^2{-}^6{C}_3{x}^3{+}^6{C}_4{x}^4{-}^6{C}_5{x}^5{+}^6{C}_6{x}^6\right)\\ =\left({ }^6{C}_0{-}^6{C}_1{x}^2{+}^6{C}_2{\left(x^2\right)}^2{-}^6{C}_3{\left(x^2\right)}^3{+}^6{C}_4{\left(x^2\right)}^4{-}^6{C}_5{\left(x^2\right)}^5{+}^6{C}_6{\left(x^2\right)}^6\right)\\ \text{ ∴The coefficient of }{x}^4{=}^6{C}_0{\cdot }^6{C}_2{-}^6{C_2}^6{C}_1{+}^6{C}_4{\cdot }^6{C}_0\\ =1\times 15-15\times 6+15\times 1\\ =30-90\\ =-60\end{array}$