Given that $\sqrt[3]{1+9x} ≈ 1+3x+ax^2+bx^3$ for small values of $x$, ﬁnd the values of the coefﬁcients $a$ and $b$ Given that $\sqrt[3]{1+9x}\approx 1+3x+ax^2+bx^3$ for small values of $x$, ﬁnd the values of the coefﬁcients $a$ and $b$. 
A question from CIE Maths Paper 33 O/N 2015. The answer is
$$a=-9$$
$$b=45$$
The mark sheet says:

State correct unsimplified $x^2$ or $x^3$ and then obtain the answers above

Any answers with explanation would be helpful.

The binomial series
$$
(1+x)^\alpha=\sum\binom\alpha k x^k
$$
gives in your case
\begin{align}
(1+9x)^{\frac13}&=1+\frac13·9x+\frac12·\frac13·(\frac13-1)·(9x)^2\\&\qquad+\frac16·\frac13·(\frac13-1)·(\frac13-2)·(9x)^3\\&\qquad+\frac1{24}·\frac13·(\frac13-1)·(\frac13-2)·(\frac13-3)·(9x)^4+...\\
&=1+3x-9x^2+45x^3-270x^4+…
\end{align}

You could also compute the third power on both sides
$$
1+9x=(1+3x)^3+3(1+3x)^2x^2(a+bx)+O(x^4)\\
=1+9x+27x^2+27x^3+3x^2(a+(6a+b)x)+O(x^4)\\
=1+9x+3(9+a)x^2+3(9+6a+b)x^3+O(x^4)
$$
