Solution:

$(x+1)^3$

A root of a polynomial $p$ will be a double root if and only if it is also a root of $p'$. Let $a$ and $b$ be the roots of $p'$. Since $a$ and $b$ are also roots of $p$, they are double roots of $p$. But $p$ can have only three roots, so $a = b$ and $a$ becomes a double root of $p'$. This makes $p'(x) = 3c(x-a)^2$ for some constant $3c$, and thus $p(x) = c(x-a)^3 + d$. Because $a$ is a root of $p$ and $p$ is monic, $d = 0$ and $c = 1$. From $p(0) = 1$ we get $p(x) = (x+1)^3$.