Let $f(x) = x^3(ax^2+bx+c)$. Then $f(x)$ is divisible by $x^3$. We want to find $a$, $b$, and $c$ such that $f(x) + 3 = x^3(ax^2+bx+c)+3$ is divisible by $(x+1)^3$. Note that $(x+1)^3 = x^3 + 3x^2 + 3x + 1$. Thus, we want to find $a$, $b$, and $c$ such that:
f(x)+3=x3(ax2+bx+c)+3=(x3+3x2+3x+1)q(x)
for some polynomial $q(x)$. Equivalently, we want to find $a$, $b$, and $c$ such that:
ax5+bx4+cx3+3=x5q(x)+3x4q(x)+3x3q(x)+q(x)
for some polynomial $q(x)$. Comparing coefficients of like powers of $x$, we get the following system of equations:
\begin{align*} a &= q(x) \ b &= 3q(x) \ c &= 3q(x) \ 3 &= q(x) \end{align*}
Thus, $q(x) = 3$, and $a = 3$, $b = 9$, $c = 9$. Therefore, the polynomial $f(x) = \boxed{3x^5 + 9x^4 + 9x^3}$ satisfies the given conditions.
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