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 $x^3(Ax^2+Bx+C)$?

We can say:
\begin{align}P(x)-1&=(x-1)^3(ax^2+bx+c)\\
P(x)&=x^3(dx^2+ex+f)
\end{align}
Therefore, we can say that \begin{align}(x-1)^3(ax^2+bx+c)+1&=x^3(dx^2+ex+f)\\
(x^3-3x^2+3x-1)(ax^2+bx+c)+1&=dx^5+ex^4+fx^3\\
ax^5+(b-3a)x^4+(c-3b+3a)x^3+(3b-3c-a)x^2+(3c-b)x+(1-c)&=dx^5+ex^4+fx^3
\end{align}
We can then equate coefficients
\begin{align}a&=d\tag{$x^5$ term}\\
b-3a&=e\tag{$x^4$ term}\\
c-3b+3a&=f\tag{$x^3$ term}\\
3b-3c-a&=0\tag{$x^2$ term}\\
3c-b&=0\tag{$x$ term}\\
1-c&=0\tag{constant term}\end{align}
We can use the last three equations to say that $c=1$, $b=3$, $a=6$
We can then use the top three equation to say that $d=6$, $e=-15$, $f=10$
We can use the equation from above, to give us:
\begin{align}P(x)&=x^3(dx^2+ex+f)\\
&=x^3(6x^2-15x+10)\\
&=6x^5-15x^4+10x^3\end{align}
