<s>[INST] 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 $x^3(Ax^2+Bx+C)$?
 [/INST] 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:
$$P(x)=6x^5-15x^4+10x^3=(6x^2+3x+1)(x-1)^3+1$$
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