Answer
The identity is true.
For proof, please see "step-by-step".
Work Step by Step
Recognize a difference of cubes in the numerator.
$x^{3}-y^{y}=(x-y)(x^{2}+xy+y^{2})$
$LHS=\displaystyle \frac{1-\cos^{3}A}{1-\cos A}=$
... substituting $x=1, y=\cos A$
$\displaystyle \frac{x^{3}-y^{3}}{x-y}=\frac{(x-y)(x^{2}+xy+y^{2})}{(x-y)}$
... reduce
$=x^{2}+xy+y^{2}$
...back substitute
$=1^{2}+\cos A+\cos^{2}A=RHS$,
proving the identity.