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 
 A: I like your idea. We can use the Pythagorean identity to simplify it to $$\sin^3 t-\cos^3 t=(\sin t-\cos t)(\sin^2 t+\cos^2 t+\sin t\cos t)=a(1+\sin t\cos t).$$
From the original equation, we know that $\sin^2 t-2\sin t \cos t+cos^2 t=1-2\sin t \cos t=a^2,$ so $$\sin t\cos t=\frac{1-a^2}{2}.$$ Now we just substitute.
A: 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 $$\cos^3(t)-\sin^3(t) = a(\cos^2(t)+\sin^2(t)+\cos(t)\sin(t))=a(1+\frac{1-a^2}{2})=\frac{a}{2}(3-a^2).$$
A: Good, you are on the right track: 
$a^3-b^3=(a-b)(a^2+ab+b^2)$
If you let $a=\cos(t)$ and $b=\sin(t)$... 
$(\cos(t))^3-(\sin(t))^3=(\cos(t)-\sin(t))((\cos(t))^2+(\cos(t))(\sin(t))+(\sin(t))^2)$
Or using the usual shorthand: 
$\cos^3(t)-\sin^3(t)=(\cos(t)-\sin(t))(\cos^2(t)+\cos(t)\sin(t)+\sin^2(t))$
Notice the Pythagorean relationship $\sin^2(t)+\cos^2(t)=1$ on the right hand side (RHS). 
And notice that $\sin(t)-\cos(t)=-(\cos(t)-\sin(t))$
Using those two facts, and the fact that $\sin(t)-\cos(t)=a$, we find that: 
$\cos^3(t)-\sin^3(t)=-(\sin(t)-\cos(t))(\sin^2(t))+\cos^2(t)+\cos(t)\sin(t)$
$\cos^3(t)-\sin^3(t)=-(a)(1+\cos(t)\sin(t))$
Now, if we know that $\sin(t)-\cos(t)=a$, then we know that $\sin^2(t)+\cos^2(t)-2\sin(t)\cos(t)=a^2$
And we know that $\frac{1-a^2}{2}=\sin(t)\cos(t)$
So, if: $\cos^3(t)-\sin^3(t)=-(a)(1+\cos(t)\sin(t))$
Then $\cos^3(t)-\sin^3(t)=-(a)(1+\frac{1-a^2}{2})$
