Given,

$$\beta = \mathop {\lim }\limits_{x \to 0} {{{e^{{x^3}}} - {{(1 - {x^3})}^{{1 \over 3}}} + ({{(1 - {x^2})}^{{1 \over 2}}} - 1)\sin x} \over {x{{\sin }^2}x}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{(1 + {x^3}\, + \,...) - \left( {1 - {{{x^3}} \over 3}\, + \,...} \right) + \left( {\left( {1 - {1 \over 2}{x^2}\, + \,...} \right) - 1} \right)x} \over {x\,.\,{{{{\sin }^2}x} \over {{x^2}}}\,.\,{x^2}}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{x^3} + {{{x^3}} \over 3} - {{{x^3}} \over 2}} \right) + {x^4}(......)} \over {{x^3}}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {1 + {1 \over 3} - {1 \over 2}} \right)} \over {{x^3}}}$$ [Neglecting higher power of x]

$$ = 1 + {1 \over 3} - {1 \over 2} = {5 \over 6}$$

$$\therefore$$ $$6\beta = 6 \times {5 \over 6} = 5$$