1. We need to evaluate

$$\beta=\lim_{x\to 0}\frac{e^{x^3}-\left(1-x^3\right)^{1/3}+\left(\left(1-x^2\right)^{1/2}-1\right)\sin x}{x\sin^2 x}.$$

2. Since $x\to 0$, use series expansions.

Step 1: Expand each term

(i) Expansion of $e^{x^3}$

$$e^{x^3}=1+x^3+\frac{x^6}{2}+\cdots$$

(ii) Expansion of $(1-x^3)^{1/3}$

Using

$$(1+u)^\alpha=1+\alpha u+\frac{\alpha(\alpha-1)}{2}u^2+\cdots$$

with $u=-x^3$, $\alpha=\frac13$,

$$(1-x^3)^{1/3}=1-\frac{x^3}{3}-\frac{x^6}{9}+\cdots$$

Therefore,

$$e^{x^3}-(1-x^3)^{1/3} =\left(1+x^3+\frac{x^6}{2}+\cdots\right)-\left(1-\frac{x^3}{3}-\frac{x^6}{9}+\cdots\right)$$ $$=\frac{4}{3}x^3+\frac{11}{18}x^6+\cdots$$

(iii) Expansion of $(1-x^2)^{1/2}-1$

Using

$$(1-x^2)^{1/2}=1-\frac{x^2}{2}-\frac{x^4}{8}+\cdots$$

so

$$(1-x^2)^{1/2}-1=-\frac{x^2}{2}-\frac{x^4}{8}+\cdots$$

Also,

$$\sin x=x-\frac{x^3}{6}+\cdots$$

Hence

$$\left((1-x^2)^{1/2}-1\right)\sin x =\left(-\frac{x^2}{2}-\frac{x^4}{8}+\cdots\right)\left(x-\frac{x^3}{6}+\cdots\right).$$

The lowest-order term is

$$-\frac{x^3}{2}+\cdots$$

So the numerator becomes

$$\frac{4}{3}x^3-\frac{1}{2}x^3+\cdots$$ $$=\left(\frac{8}{6}-\frac{3}{6}\right)x^3+\cdots=\frac{5}{6}x^3+\cdots$$

3. Now expand the denominator:

$$x\sin^2 x=x\left(x-\frac{x^3}{6}+\cdots\right)^2.$$

Since

$$\sin^2 x=x^2+O(x^4),$$

we get

$$x\sin^2 x=x^3+O(x^5).$$

4. Therefore,

$$\beta=\lim_{x\to 0}\frac{\frac{5}{6}x^3+\cdots}{x^3+\cdots}=\frac{5}{6}.$$

Hence,

$$6\beta=6\cdot \frac{5}{6}=5.$$

Final Answer

$$\boxed{5}$$

The derived answer matches the stored correct answer.