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 $x^3(Ax^2+Bx+C)$?
 A: 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 $P(x)=\frac{30}{5}x^5-\frac{30}{2}x^4+\frac{30}{3}x^3=6x^5-15x^4+10x^3$.
A: Expanding $(x-1)^3$ we find that
$$P(x)=(Ax^2+Bx+C)(x^3-3x^2+3x-1)+1$$
$$=Ax^5+(B-3A)x^4+(C-3B+3A)x^3+(-3C+3B-A)x^2+(3C-B)x-C+1$$
Yet we also know that $P(x)$ is divisible by $x^3$ and is of the form
$$Dx^5+Ex^4+Fx^3$$
Comparing constant, linear and quadratic coefficients we have
$$1-C=0$$
$$3C-B=0$$
$$-3C+3B-A=0$$
and from these we get $C=1$, $B=3$ and $A=6$ in order. The remaining coefficients, $x^3$ to $x^5$, give us the coefficients of $P(x)$:
$$D=A=6$$
$$E=B-3A=-15$$
$$F=C-3B+3A=10$$
In conclusion:
$$P(x)=6x^5-15x^4+10x^3=(6x^2+3x+1)(x-1)^3+1$$
A: 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}
A: $P(x)=1+(ax^2+bx+c)(x-1)^3=1-(c+bx+ax^2)(1-3x+3x^2-x^3)$
$=1-c+x(3c-b)+x^2(-a+3b-3c)+x^3(\cdots)$
We need $1-c=3c-b=-a+3b-3c=0$   
