Number of integral values of b for which the equation $\frac{x^3}{3}-x=b$  has three distinct solution is____
 

$f\left(x\right)=\frac{x^3}{3}-x-6$

$\therefore f\left(x\right)=x^2-1=0\therefore x=1\text{ or }-1$

For three distinct roots, $f\left(x_1\right)f\left(x_2\right)<0$

where $x_1$ and $x_2$ are the roots of $f^{\mathrm{\prime }}\left(x\right)=0.$ Thus,

$\left(\frac{1}{3}-1-b\right)\left(-\frac{1}{3}+1-b\right)<0\text{ or }\left(b+\frac{2}{3}\right)\left(b-\frac{2}{3}\right)<0$

$\text{ or }b\in \left(-\frac{2}{3},\frac{2}{3}\right)$