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

The correct answer is 5.

We are given the limit:
$\beta =\underset{x\to 0}{lim}\frac{e^{x^3}-{(1-x^3)}^{\frac{1}{3}}+({(1-x^2)}^{\frac{1}{2}}-1)sinx}{x{sin}^{2}x}$

Step 1: Simplify the denominator using standard limits
Recall that $\underset{x\to 0}{lim}\frac{sinx}{x}=1$.
Therefore, ${sin}^{2}x$ behaves like $x^2$ as $x\to 0$.
The denominator simplifies to:
$x{sin}^{2}x=x^3{\left(\frac{sinx}{x}\right)}^2\approx x^3$

Step 2: Expand the terms in the numerator up to order $x^3$
Using Taylor series expansions around $x=0$:

1. Expansion of $e^{x^3}$:
$e^{x^3}=1+x^3+O\left(x^6\right)$

2. Expansion of ${(1-x^3)}^{\frac{1}{3}}$ using the binomial expansion ${(1+y)}^n=1+ny+\dots$:
${(1-x^3)}^{\frac{1}{3}}=1-\frac{1}{3}{x}^3+O\left(x^6\right)$

3. Expansion of the third term $({(1-x^2)}^{\frac{1}{2}}-1)sinx$:
Using binomial expansion:
${(1-x^2)}^{\frac{1}{2}}=1-\frac{1}{2}{x}^2+O\left(x^4\right)$
So,
${(1-x^2)}^{\frac{1}{2}}-1=-\frac{1}{2}{x}^2+O\left(x^4\right)$
Since $sinx=x-\frac{x^3}{6}+O\left(x^5\right)$, we have:
$({(1-x^2)}^{\frac{1}{2}}-1)sinx=(-\frac{1}{2}{x}^2+O\left(x^4\right))(x+O\left(x^3\right))=-\frac{1}{2}{x}^3+O\left(x^5\right)$

Step 3: Combine all terms in the numerator
Numerator $=(1+x^3)-(1-\frac{1}{3}{x}^3)+(-\frac{1}{2}{x}^3)+O\left(x^4\right)$
Numerator $=1+x^3-1+\frac{1}{3}$
Numerator $=(1+\frac{1}{3}-\frac{1}{2})x^3+O\left(x^4\right)$
Numerator $=\left(\frac{6+2-3}{6}\right)x^3+O\left(x^4\right)=\frac{5}{6}{x}^3+O\left(x^4\right)$

Step 4: Evaluate the limit $\beta$
$\beta =\underset{x\to 0}{lim}\frac{\frac{5}{6}{x}^3+O\left(x^4\right)}{x^3}=\frac{5}{6}$

Step 5: Find the value of $6\beta$
$6\beta =6\times \frac{5}{6}=5$

Thus, the value of $6\beta$ is 5.