Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$ 
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.

No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?

Less bashy than Parcly Taxel's answer:
Since $P$ is divisible by $x^3$ and $P-1$ is divisible by $(x-1)^3$, we know that $P'$ is divisible by $x^2(x-1)^2$.  Since $P'$ is degree $4$, it must be a constant multiple of this.  Say $P'(x)=Ax^2(x-1)^2=Ax^4-2Ax^3+Ax^2$.
Then $P(x)$ is an antiderivative of this, namely $\frac{A}{5}x^5-\frac{A}{2}x^4+\frac{A}{3}x^3+B$.  At $0$ this is $0$ and at $1$ this is $1$, so we get $B=0$ and $\frac{A}{30}=1$.  Thus $P(x)=\frac{30}{5}x^5-\frac{30}{2}x^4+\frac{30}{3}x^3=6x^5-15x^4+10x^3$.
