To solve this problem let's multiply (sin x - cos x) and (1 + sin x* cos x).

(sin x - cos x)*(1 + sin x*cos x)

= sin x + (sin x)^2 * cos x - cos x - sin x * (cos x )^2

Now (sin x)^2 + (cos x)^2 = 1 or (sin x)^2 = 1- (cos x)^2 and (cos x)^2 = 1 - (sin x)^2.

So sin x + (sin x)^2 * cos x - cos x - sin x * (cos x )^2

= sin x + [1- (cos x)^2] * cos x - cos x - sin x *[ 1 - (sin x)^2]

= sin x + cos x - (cos x)^3 - cos x - sin x + ( sin x)^3

subtracting common terms

= ( sin x)^3 - (cos x)^3

**Therefore sin^3 x - cos^3 x = (sin x - cos x)(1 + sin x*cos x)**

To prove sin^3 x-cos^3x = (sinx-cosx)(1+sinx+cosx).

We use the identity a^3-b^3 = (a-b)(a^2+ab+b^2).

and the trigonometric identity sin^2x+cos^2x = 1 to prove the given enunciation.

Therefore using the first identity,

sin^3x -cos^3x = (sinx-cosx){ sin^2x+sinx*cosx+cos^2x)

sin^3x-cos^3x = (sinx-cosx)[(sin^2x+cos^2x)+ sinxcosx}

sin^3x-cos^3x = (sinx -cosx)(1 +sinxcosx) , using the second identity above.

We'll have to verify if the expression from the left side is equal to the expression from the right side.

We'll start by re-writting the difference of the cubes from the left side. We'll use the formula:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

We'll substitute a and b by sin x and cos x and we'll get:

(sin x)^3 - (cos x)^3 = (sin x - cos x)[(sin x)^2 + sin x*cos x + (cos x)^2]

But the sum (sin x)^2 + (cos x)^2 = 1, from the fundamental formula of trigonometry.

We'll substitute the sum of squares by the value 1.

**(sin x)^3 - (cos x)^3 = (sin x - cos x)(1 + sin x*cos x) q.e.d**