If $\int x^3{e}^{-x}dx=-e^{-x}[a{x}^3+b{x}^2+cx+d]+K$ then (a, b, c, d)

$\int x^3{e}^{-x}dx$

$=x^3\left(\frac{{\displaystyle e^{-x}}}{{\displaystyle -1}}\right)-\int \left(3x^2\right)\left(\frac{e^{-x}}{-1}\right)dx$

$=-x^3{e}^{-x}+3\int x^2{e}^{-x}dx$

$=-x^3{e}^{-x}+3\left[x^2\left(\frac{{\displaystyle e^{-x}}}{{\displaystyle -1}}\right)-\int 2x(-e^{-x})dx\right]$

$=-x^3{e}^{-x}-3x^2{e}^{-x}+6\int x.e^{-x}dx$

$=-x^3{e}^{-x}-3x^2{e}^{-x}+6\left[x(-e^{-x})-\int 1(-e^{-x})\right]$

$=-x^3{e}^{-x}-3x^2{e}^{-x}-6x{e}^{-x}+6(-e^{-x})+c$

$=-e^{-x}[x^3+3x^2+6x+6]+c$

$(a,b,c,d)=(1,3,6,6)$