Answer
The identity is verified.
$cos^3x~sin^2x=(sin^2x-sin^4x)~cos~x$
Work Step by Step
$sin^2x+cos^2x=1$ (Pythagorean identity)
$cos^2x=1-sin^2x$
$cos^3x~sin^2x=cos~x~cos^2x~sin^2x=cos~x~(1-sin^2x)~sin^2x=cos~x~(sin^2x-sin^4x)=(sin^2x-sin^4x)~cos~x$
