If $f(x) = sin^3(x) + cos^3(x)$ prove that $f''(x)$ If $$f(x) = \sin^3(x) + \cos^3(x),$$ prove that $$f''(x) = \frac 32 [\sin(x) + \cos(x)] [3\sin(2x) - 2]$$
I had my go at this and I ended up with $$\frac32[\sin(x)+\cos(x)] [4\sin(2x) - 2]$$ which is a wrong answer.
Note that $\sin^3x+\cos^3x=( \sin x)^3+(\cos x)^3$ so you can use the chain rule to get $$f'(x)=3(\sin x)^2\cos x-3(\cos x )^2\sin x= 3\sin x \cos x(\sin x -\cos x).$$ Then use the product rule 
$$f''(x)=3(\cos^2x-\sin^2x)(\sin x-\cos x)+3\sin x \cos x(\cos x+\sin x)$$ Which you can simplify however you want. One way would be to use the identity $2\sin x \cos x $ to write $3\sin x\cos x= \frac 32\sin(2x)$ and the identites $\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x$.
http://www.derivative-calculator.net/

This does it for you and also provides many many steps