$\text{ The sum of the cubes of the coefficients in the expansion of }(x-y{)}^3\text{ is }$

${(x-y)}^n={}^{n}C_0{x}^n{y}^0-{}^{n}C_1{x}^{n-1}{y}^1+{}^{n}C_2{x}^{n-2}{y}^2-\mathrm{........}+{(-1)}^n{}^{n}C_n{x}^0{y}^n$

$\therefore {(x-y)}^3={}^{3}C_0{x}^3{y}^0-{}^{3}C_1{x}^2{y}^1+{}^{3}C_2{x}^1{y}^2-{}^{3}C_3{x}^0{y}^3$

$=x^3-3x^{2}y+3x{y}^2-y^3$

The coefficients are 1,-3,3,-1

Sum of cubes is $1^3+{(-3)}^3+{(+3)}^3+{(-1)}^3$

$=1-27+27-1$

= 0