Answer
The identity is true.
For details, please see "step-by-step".
Work Step by Step
Recognize a difference of cubes in the numerator.
$x^{3}-y^{y}=(x-y)(x^{2}+xy+y^{2})$
$LHS=\displaystyle \frac{\sin^{3}A-8}{\sin A-2}=$
... substituting $x=\sin A, y=2$
$\displaystyle \frac{x^{3}-y^{3}}{x-y}=\frac{(x-y)(x^{2}+xy+y^{2})}{(x-y)}$
$=x^{2}+xy+y^{2}$
...back substitute
$=\sin^{2}A+2\sin A+2^{2}=RHS$,
proving the identity.