Prove that sin^3x=sinx+sin^2x*cosx-sinx8cos^2x-cosx+cos^3x.

We'll subtract (cos x)^3 both sides to get a difference of
cubes to the left side:

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

We'll factorize the terms
from the middle by sinx*cos x:

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

We'll factorize by sin
x - cos x, to the left side:

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

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