Answer
$(6,-3,5,1)$
Work Step by Step
We know that $a(x^3+x^2)+b(x^3-1)+c(x^3+1)+d(x^3+x)=8+x+6x^2+9x^3$.
Thus $ax^3+ax^2+bx^3-b+cx^3+c+dx^3+dx\\
=(-b+c)+dx+ax^2+(a+b+c+d)x^3\\
=8+x+6x^2+9x^3$
Thus $-b+c=8\\ d=1 \\ a=6 \\ a+b+c+d=9$
Thus $2c=10 \rightarrow c=5\\ b+5=2$
Thus $a=6 \\ b =- 3 \\ c=5 \\ d=1$
Thus the vector is $(6,-3,5,1)$.
