<s>[INST] Find Maclaurin series of $(\sin(x^3))^{1/3}$ How do I find Maclaurin series for the function:
$$\sqrt[3]{\sin(x^3)}$$
The answer should be:
$$ x - \frac {x^7}{18} - \frac {{x}^{13}}{3240} + o(x^{13})$$
I tried:
$$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} -  \frac {x^7}{7!} + ...$$
So, I changed $x$ to $x^3$ to get:
$$\sin(x^3) = x^3 - \frac {x^9}{3!} + \frac {x^{15}}{5!} -  \frac {x^{21}}{7!} + ...$$
But, I'm stuck when it comes to power of 1/3:
$$\sqrt[3]{x^3 - \frac {x^9}{3!} + \frac {x^{15}}{5!} -  \frac {x^{21}}{7!} + ...} = a_0+a_1x+a_2x^2+a_3x^3+...$$
 [/INST] If you factor out $x^3$ from cubic root you'll get
$$
x\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )}
$$
Now, use power series expansion for cubic root
$$
\sqrt[3]{1-x} = 1-\frac x3-\frac {x^2}9+o(x^2) \\
\sqrt[3]{1-\left (\frac{x^6}{3!}-\frac{x^{12}}{5!} +o(x^{12})\right )} = 1-\frac 13\left(\frac{x^6}{3!}-\frac{x^{12}}{5!} \right)-\frac 19\left(\frac {x^{12}}{3!3!}+o(x^{12}) \right ) = \\
\\= 1-\frac {x^6}{3\cdot 3!}+\left ( \frac 1{3\cdot5!}-\frac 1{9\cdot 3! \cdot 3!}\right)x^{12}+o(x^{12}) = 1-\frac {x^6}{18}-\frac {x^{12}}{3240}+o(x^{12})
$$
After multiplying to $x$ you'll get your answer.
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