<s>[INST] 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 $x-x^2$??
 [/INST] Note that one has $o(x^6)=o(x^5)$ (as $x\to 0$), but not $o(x^5)=o(x^6)$. 
Also, $o((x-x^2)^2)=o(x^5)$ is incorrect: one has $\lim_{x\to 0}\frac{x^5}{(x-x^2)^2}=0$ but $\lim_{x\to 0}\frac{x^5}{x^5}=1$. In general, $o(f(x))=o(g(x))$ does NOT imply that $o(g(x))=o(f(x))$ as discussed in an answer to another question of yours.

Why the answer cannot be $x-x^2$?

It is an instructive exercise to show (using the correct answer in Apostol) that the following is FALSE:
$$
\lim_{x\to 0}\frac{\sin (x-x^2)-(x-x^2)}{x^6}= 0\tag{1}
$$

[Added.]
To see why (1) is false, you can use the solution of $P(x)$ in the book
so that
$$\begin{align}
f(x):=\frac{\sin (x-x^2)-(x-x^2)}{x^6}
&=\frac{a_3x^3+a_4x^4+a_5x^5+a_6x^6+o(x^6)}{x^6}\\
&=\frac{1}{x^3}
\left(
a_3+a_4x+a_5x^2+a_6x^3
\right)+\frac{o(x^6)}{x^6}\\
&=\frac{1}{x^3}\cdot g(x)+h(x)
\end{align}
$$
where $g(x):=a_3+a_4x+a_5x^2+a_6x^3$.
Now, note that
$$
\lim_{x\to 0}g(x)=a_3\neq 0,\quad \lim_{x\to 0}\frac{o(x^6)}{x^6}=0.
$$
So
$
\lim_{x\to 0+}f(x)=\infty
$
and (1) is thus impossible.
</s>