Purdue Problem of the week, Fall 2011 Problem 4

The problem 4 of 2011 fall is:

 Show that if: u(x)= Sum { x^(3k)}/{(3k)!};  

                     v(x)= Sum {x^(3k+1)}/{(3k+1)!};

                     w(x)=Sum {x^(3k+2)}/{(3k+2)!}; where k=0,1,2,3,...

then u^3+v^3+w^3-2uvw=1.

Let f(x) = u^3+v^3+w^3-2uvw, then consider f'(x).

f'(x)=3u'u^2+3v'v^2+3w'w^2-3u'vw-3uv'w-3uvw'.

It is easy to calculate u'=w, v'=u, w'=v. Therefore f'(x)=0,  that means f(x) is constant.

Because f(0)=1, so f(x)=1 for all value of x.