If $\beta = \lim_{x \to 0} \frac{e^{x^{3}} - \left(1 - x^{3}\right)^{\frac{1}{3}} + \left(\left(1 - x^{2}\right)^{\frac{1}{2}} - 1\right)\sin x}{x \sin^{2} x}$, then the value of $6\beta$ is ______.

If β = lim_(x → 0) (e^(x³) - (1 - x³)^(1/3) + ((1 - x²)^(1/2) - 1)sin x)/(x sin² x), then the value of 6β is ______. $\beta = \lim_{x \to 0} \dfrac{e^{x^{3}} - (1 - x^{3})^{1/3} + \left((1 - x^{2})^{1/2} - 1\right)\sin x}{x \sin^{2} x}$ $\text{At } x \to 0:\ \text{numerator} \to 0,\ \text{denominator} \to 0 \Rightarrow \dfrac{0}{0} \text{ form}$ $\text{Find } \beta, \text{ then } 6\beta$ $\text{Plan: expand numerator \& denominator in powers of } x,\ \text{keep leading terms}$ $\sin x = x - \dfrac{x^3}{6} + \cdots \Rightarrow \sin^2 x \approx x^2$ $x \sin^2 x \approx x \cdot x^2 = x^3$ $e^{x^3} = 1 + x^3 + \dfrac{x^6}{2} + \cdots$ $(1 - x^3)^{1/3} = 1 - \dfrac{x^3}{3} - \dfrac{x^6}{9} + \cdots$ $e^{x^3} - (1 - x^3)^{1/3} = x^3 + \dfrac{x^3}{3} + \cdots = \dfrac{4}{3}x^3 + \cdots$ $(1 - x^2)^{1/2} - 1 = -\dfrac{x^2}{2} + \cdots, \qquad \sin x \approx x$ $\left((1 - x^2)^{1/2} - 1\right)\sin x = -\dfrac{x^3}{2} + \cdots$ $\text{num} = \dfrac{4}{3}x^3 - \dfrac{1}{2}x^3 + \cdots$ $= \left(\dfrac{4}{3} - \dfrac{1}{2}\right)x^3 = \dfrac{5}{6}x^3 + \cdots$ $\beta = \lim_{x \to 0} \dfrac{\tfrac{5}{6}x^3 + \cdots}{x^3 + \cdots} = \dfrac{5}{6}$ $6\beta = 6 \cdot \dfrac{5}{6} = 5$ $\boxed{6\beta = 5}$ $\beta = \dfrac{5}{6}, \qquad 6\beta = 5$