Given:

The given equations are as follows.

  u=1+x33!+x66!+x99!+⋯v=x+x44!+x77!+x1010!+⋯w=x22!+x55!+x88!+⋯

Calculation:

Differentiate the equation u=1+x33!+x66!+x99!+⋯ with respect to x .

  dudx=ddx(1+x33!+x66!+x99!+⋯)u'=3x23!+6x56!+9x89!+⋯=x22!+x55!+x88!+⋯=w

Differentiate the equation v=x+x44!+x77!+x1010!+⋯ with respect to x .

  dvdx=ddx(x+x44!+x77!+x1010!+⋯)v'=1+x33!+x66!+x99!+⋯=u

Differentiate the equation w=x22!+x55!+x88!+⋯ with respect to x .

  dwdx=ddx(x22!+x55!+x88!+⋯)w'=x+x44!+x77!+x1010!+⋯=v

In order to show that u3+v3+w3−3uvw=C , take the left hand side term and differentiate.

Differentiate the left hand side term with respect to x .

  ddx(u3+v3+w3−3uvw)=3u2u'+3v2v'+3w2w'−3(u'vw+uv'w+uvw')=3u2w+3v2u+3w2v−3(vw2+u2w+uv2)=0

Integrate the equation on both sides.

  ∫ddx(u3+v3+w3−3uvw)dx=∫0dxu3+v3+w3−3uvw=C

Substitute 0 for x in the equation given equations to find the values of u , v , and w .

  u=1v=0w=0

Substitute 1 for u , 0 for v , and 0 for w in the equation u3+v3+w3−3uvw=C to find the value of C .

  (1)3+(0)3+(0)3−3(1)(0)(0)=CC=1

Substitute 1 for C in the equation u3+v3+w3−3uvw=C .

  u3+v3+w3−3uvw=1

Therefore, it is proved that u3+v3+w3−3uvw=1 .