Using trigonometric identities

**sinx - sinx cos ^{2}x = sin^{3}x**

*remember that* **sin ^{2}x + cos^{2}x = 1**,

*so*

**cos**,

^{2}x = 1 - sin^{2}x*and we can substitute that in place of*

**cos**...

^{2}x**sinx - sinx (1 - sin ^{2}x) = sin^{3}x**

*now distribute the*

**sinx**

*into the parentheses*...

**sinx - (sinx - sin ^{3}x) = sin^{3}x**

**sinx - sinx + sin ^{3}x = sin^{3}x**

**sin ^{3}x = sin^{3}x**

## Comments

are you sure about the problem? sinx-sinxcos^2x is eqypual to sin^3x.

Oops...my bad! Your problem is correct! sinx-sinxcos^2x is eqypual to sin^3x

How can I verify the identity?