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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
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