Proving a trigonometric equation Knowing that :    $$ \sin t - \cos t = x $$   
Prove that :    $$ \cos^3t - \sin^3t = \frac{x}{2}(3-x^2)  $$     
I tried to solve it by the important corresponding $$ a^3 - b^3 = (a-b)(a²+ab+b²) $$ 
But I got stuck in the middle and I don't even know if it's correct what I did 

Note that $$a^2=(\sin(t)-\cos(t))^2=\sin^2(t)+\cos^2(t)-2\sin(t)\cos(t)=1-2\sin(t)\cos(t).$$ Hence $$\cos^3(t)-\sin^3(t) = a(\cos^2(t)+\sin^2(t)+\cos(t)\sin(t))=a(1+\frac{1-a^2}{2})=\frac{a}{2}(3-a^2).$$
