Question

Another question buried in the back of my mind:
If \[u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots\\
v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\cdots\\
w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\cdots\]
prove that \(\Large u^3+v^3+w^3-3uvw=1\).
Of course, I know that u+v+w=e^x, but I don't know how that helps.
Also, I know that \(\large u^3+v^3+w^3-3uwv=(u+v+w)(u^2+v^2+w^2-uv-uw-vw)\) and again, I don't know how this helps.

Answer 1

\[\frac{1}{2}\left(2u^2 + 2v^2 + 2w^2 -2uv -2uw -2vw\right) = \frac{1}2\left((u-w)^2 + (w-v)^2 + (v-u)^2\right)\]

Answer 2

I don't think it's gonna help :(

Answer 3

I wonder if it can magically be proven that lshaan's expression is equal to e^-x.

Answer 4

it is a little bit tricky
find the Derivative of u3+v3+w3−3uvw and it will = 0 and u can infer the u3+v3+w3−3uvw=constant because the Derivative of a constant is zero then put x=0 and u will conclude that the constant is one

Answer 5

@RED123url Why didn't I think of that? -_-
Noticing that u'=w, v'=u, and w'=v:
\[(u^3+v^3+w^3-3uvw)'=3u^2u'+3v^2v'+3w^2w'-3u'vw-3uv'w-3uvw'\\
=3u^2w+3uv^2+3w^2v-3w^2v-3u^2w-3uv^2=0\]
And from here, I know how to proceed.
Thanks so much for helping me! I can finally lay one of my demons to rest.

Answer 6

u r welcome