Simplify $\frac{{\sin }^{3}A+{\cos }^{3}A}{\sin A+\cos A}$

The correct answer is:
$1-\sin A\cos A$

To simplify the given expression, let us recall the algebraic identity for the sum of two cubes:
$x^3+y^3=(x+y)(x^2-xy+y^2)$

In the numerator of the given expression, we have the sum of two cubes where $x=\sin A$ and $y=\cos A$.

Applying the sum of cubes identity to the numerator:
${\sin }^{3}A+{\cos }^{3}A=(\sin A+\cos A)({\sin }^{2}A-\sin A\cos A+{\cos }^{2}A)$

Now, substitute this factored form back into the original expression:
$\frac{{\sin }^{3}A+{\cos }^{3}A}{\sin A+\cos A}=\frac{(\sin A+\cos A)({\sin }^{2}A-\sin A\cos A+{\cos }^{2}A)}{\sin A+\cos A}$

Since $\sin A+\cos A\ne 0$, we can cancel out the common factor $(\sin A+\cos A)$ from both the numerator and the denominator:
$={\sin }^2-\sin A\cos A+{\cos }^{2}A$

Using the fundamental trigonometric identity ${\sin }^{2}A+{\cos }^{2}A=1$, we can group the terms:
$=({\sin }^2+{\cos }^2)-\sin A\cos A$

Substituting 1 for ${\sin }^{2}A+{\cos }^{2}A$ gives the simplified expression:
$=1-\sin A\cos A$