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

$\beta =\underset{x\to 0}{\lim }\frac{e^{x^3}-{\left(1-x^3\right)}^{\frac{1}{3}}+\left({\left(1-x^2\right)}^{\frac{1}{2}}-1\right)\sin x}{x·\frac{{\sin }^{2}x}{x^2}·x^2}$

$\beta =\underset{x\to 0}{\lim }\left(\frac{\left(1+x^3+\frac{x^6}{2!}+\cdots \right)-\left(1-\frac{1}{3}{x}^3+\left(-\frac{1}{9}\right)x^6+\cdots \right)+\left(-\frac{1}{2}{x}^2-\frac{1}{8}{x}^4+\cdots \right)\left(x-\frac{x^3}{3!}+\cdots \right)}{x^3}\right)$

$\beta =\underset{x\to 0}{\lim }\frac{x^3\left(1+\frac{1}{3}-\frac{1}{2}\right)}{x^3}$  (Neglecting higher powers of $x$)

$\beta =\frac{5}{6}$ 

i.e. $6\beta =5$