Answer
$$\sin^3x-\cos^3x=(\sin x-\cos x)(1+\sin x\cos x)$$
Work Step by Step
$$A=\sin^3x-\cos^3x$$
Now it is crucial here not to forget that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
which means
$$A=(\sin x-\cos x)(\sin^2 x+\sin x\cos x+\cos^2 x)$$
$$A=(\sin x-\cos x)[(\sin^2 x+\cos^2 x)+\sin x\cos x]$$
- From Pythagorean Identity: $$\sin^2 x+\cos^2x=1$$
So, $A$ would be $$A=(\sin x-\cos x)(1+\sin x\cos x)$$
