First check the identity: let x=π/4, then sin3(x)+cos3(x)=2sin3(x)=2(1/√2)3=2/(2√2)=1/√2.
sin(x)cos(x)=½, so (sin3(x)+cos3(x))/(sin(x)cos(x))=√2. 1-½sin(2x)=1-½=½, so the identity is false and therefore cannot be proved. Here's what I think you meant:
½sin(2x)=sin(x)cos(x).
sin3(x)+cos3(x)=(sin(x)+cos(x))(sin2(x)-sin(x)cos(x)+cos2(x)),
sin3(x)+cos3(x)=(sin(x)+cos(x))(1-½sin(2x)),
(sin3(x)+cos3(x))/(sin(x)+cos(x))=1-½sin(2x).
Test the proposed identity: x=π/4: sin(x)+cos(x)=2sin(x)=√2; (1/√2)/√2=½; 1-½sin(2x)=½, so identity is true for x=π/4. Now let x=0: 1/1=1, also true.