己知恒等式x^3+x^2+x+1=a+b(x+1)+c(x+1)(x+3)+d(x+1)(x+3)(x+5),试求a,b,c,d的值
问题描述：
己知恒等式x^3+x^2+x+1=a+b(x+1)+c(x+1)(x+3)+d(x+1)(x+3)(x+5),试求a,b,c,d的值
答
a+b(x+1)+c(x+1)(x+3)+d(x+1)(x+3)(x+5)=a+(bx+b)+(cx^2+4cx+3c)+(dx^3+9dx^2+23dx+15d)=(a+b+3c+15d)+(b+4c+23d)x+(c+9d)x^2+dx^3=x^3+x^2+x+1得：d=1c+9d=1b+4c+23d=1a+b+3c+15d=1 解得a=0,b=10,c=-8,d=1