The sum of the coefficients in the binomial expansion  ${(a-x)}^5$

$$\begin{gathered} {(a-x)}^n={}^{n}C_0{a}^n{x}^0-{}^{n}C_1{a}^{n-1}{x}^1+{}^{n}C_2{a}^{n-2}{x}^2-\mathrm{........}+{(-1)}^n{}^{n}C_n{a}^0{x}^n \\ \therefore {(a-x)}^5={}^{5}C_0{a}^5{x}^0-{}^{5}C_1{a}^4{x}^1+{}^{5}C_2{a}^3{x}^2-{}^{5}C_3{a}^2{x}^3+{}^{5}C_4{a}^1{x}^4-{}^{5}C_5{a}^0{x}^5 \\ =a^5-5a^{4}x+10a^3{x}^2-10a^2{x}^3+5a{x}^4-x^5 \\ According to problem sum of coefficients are \\ 1-5+10-10+5-1=0 \end{gathered}$$