Why is $x^x-(\sin x)^x\sim\frac{1}{6}x^3 $ when $x\to 0$? I'm learning Taylor's expansion.
The given solution of the problem is:

when $x\to 0$, $x^x-(\sin x)^x=x^x(1-(\frac{\sin x}{x})^x)\sim\frac{1}{6}x^3 $

But I don't know how to use Taylor's formula to get:

$x^x(1-(\frac{\sin x}{x})^x)\sim\frac{1}{6}x^3 $


Just use
$$\sin(x) \approx x - \frac{x^3}{6}$$
$$\frac{\sin x}{x} \approx 1 - \frac{x^2}{6}$$
$$x^x\left(1 - \left(1 - \frac{x^2}{6}\right)^x\right)$$
Now the bracket factor
$$\left(1 - \frac{x^2}{6}\right)^x$$
is small, so you can use Binomial expansion: $(1+X)^a = 1 + aX + \ldots$
That is
$$x^x\left(1 - \left( 1 - x\cdot \frac{x^2}{6}\right)\right) = x^x\left(1 - 1 + \frac{x^3}{6}\right) = x^x\left(\frac{x^3}{6}\right)$$
For $x\to 0$ $x^x \to 1$ hence you remain with
$$\frac{x^3}{6}$$
