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 $x-x^2$??

What they are asking you is to find a polynomial $p(x)$ such that
$$
\lim_{x\rightarrow0}\frac{|\sin(x-x^2)-p(x)|}{x^6}=0
$$
No to your statements: 


*

*$o((x-x^2)^2) =o(x^5)$ is in general false. For example $h(x)=x(x-x^2)^2=o((x-x^2)^2$ as $x\rightarrow0$ but $x^{-5}h(x)\nrightarrow0$ as $x\rightarrow0$. 

*$o(x^5) = o(x^6)$ is false in general: $g(x):=x^6=o(x^5)$ but $g(x) \neq o(x^6)$.


To help you solve the problem, use Taylor expansion of $\sin z= z-\frac{z^3}{6}+\frac{z^5}{5!}-\frac{z^7}{7!}+...$. Notice that when $z=x-x^2$, $z^7$ is a polynomial on $x$ of degree 14 of the $z^7=x^7-7x^8+...-x^{14}$ This term is of course $o(x^6)$ since all powers involved are larger than $6$. So truncating the sine series up to the $7-th$ power gives the right decay of residual. The rest is to check that the 
$$
(x-x^2)-\frac{(x-x^2)^3}{6}+\frac{(x-x^2)^5}{5!}
$$
is the polynomial in the statement of your problem. A little but tedious algebra.
