The coefficient of $x^{10}$ in the expansion of ${\left(1+x^2-x^3\right)}^8,$ is

Given expansion can be rewritten as

${\left[1+x^2(1-x)\right]}^8$

$$\begin{gathered} ={}^{8}C_0+{}^{8}C_1{x}^2(1-x)+{}^{8}C_2{x}^4(1-x{)}^2 \\ +{}^{8}C_3{x}^6(1-x{)}^3+{}^{8}C_4{x}^8(1-x{)}^4+{}^{8}C_5{x}^{10}(1-x{)}^5+\cdots \end{gathered}$$

There are only two terms, which we get the coefficient of $x^{10}$.

$\therefore$ Coefficient of $x^{10}={}^{8}C_4$[Coefficient of $x^2$ in $(1-x{)}^4$]

$$\begin{gathered} and {}^{8}C_5\text{ [Coefficient of }{x}^0\mathrm{in}(1-x{)}^5] \\ ={}^{8}C_4\left({}^{4}C_2\right)+{}^{8}C_5\left(1\right) \\ =\left({}^{8}C_4\right)\left({}^{4}C_2\right)+{ }^8{C}_3 \\ =\left(70\right)\left(6\right)+56 \\ =476 \end{gathered}$$