Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:

Find the polynomial $P(x)$ of minimal degree such that
  $$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$

and the answer in the book is

$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}
$$

I am wondering why the answer cannot be 
$P(x) = x-x^2$
Since we can use the fact that $\sin(x-x^2) = x-x^2 + o((x-x^2)^2)$ as $x \to 0$
And we know that $o((x-x^2)^2) = o(x^5) = o(x^6)$ as $x$ goes to zero.
Then why is the answer from the back of the book not equal to $x-x^2$??

A sketch of the correct method:
Since $x-x^2$ has order $1$, you have to expand $\sin u$ up to degree $6$, and substitute $x-x^2$ in $u, u^2,\dots u^6$, truncating the expansions of the different $(x-x^2)^k$ at degree $6$, which leads to a recursive computation:
\begin{align}
(x-x^2)^2&=x^2-2x^3+x^4& \quad(x-x^2)^3&=x^3-3x^4+3x^5-x^6 \\
(x-x^2)^4&=x^4-4x^5+6x^6-4x^7+x^8&(x-x^2)^5&=(x-x^2)^4(x-x^2)\\
&=x^4-4x^5+6x^6+o(x^6)&&=\bigl((x^4-4x^5+6x^6+o(x^6)\bigr)(x-x^2)\\
(x-x^2)^6&=(x-x^2)^5(x-x^2)&&=\begin{aligned}[t]&x^5-4x^6+6x^7+o(x^7)\\&\phantom{x5}-x^6+4x^7-6x^8+o(x^8)
&\end{aligned}\\&=\bigl(x^5-5x^6+o(x^6)\bigr)(x-x^2)&&=x^5-5x^6+o(x^6)\\
&=\dots=x^6+o(x^6)
\end{align}
Can you end the computations?
