Challenge!

(1)
If a≠b and a≠-b,
(3a² + b²) (x² - x + 1) = (3b² + a²) (x² + x + 1)
(3a² + b²) x² - (3a² + b²) x + (3a² + b²) = (3b² + a²) x² + (3b² + a²) x + (3b² + a²)
(2a² - 2b²) x² - (4a² + 4b²) x + (2a² - 2b²) = 0
(a² - b²) x² - (2a² + 2b²) x + (a² - b²) = 0
(a - b) (a + b) x² - (2a² + 2b²) x + (a - b) (a + b) = 0
[(a - b) x - (a + b)] [(a + b) x - (a - b)] = 0
x = (a+b)/(a-b) or x = (a-b)/(a+b)

If a=b≠0,
The original equation becomes
4a² (x² - x + 1) = 4a² (x² + x + 1)
8a² x = 0
x = 0

If a=b=0,
The equation becomes 0 = 0 and so every real x satisfies the equation.

If a=-b≠0,
The original equation becomes
4a² (x² - x + 1) = 4a² (x² + x + 1)
8a² x = 0
x = 0


(2)
In order for easy typing, let u = x1/3, v = y1/3, w = z1/3,
Then the question becomes
"If u+v+w=0, show that (u³ + v³ + w³)³ = 27 u³ v³ w³"

Since u+v+w=0, by taking cube on both sides,
(u+v+w)³ = 0
u³ + v³ + w³ + 3u²v + 3uv² + 3v²w + 3vw² + 3u²w + 3uw² + 6uvw = 0【Check by yourself】
(u³ + v³ + w³) + (3u²v + 3uv²) + (3v²w + 3vw²) + (3u²w + 3uw²) + 6uvw = 0
(u³ + v³ + w³) + (3u²v + 3uv² + 3uvw) + (3v²w + 3vw² + 3uvw) + (3u²w + 3uw² + 3uvw) + 6uvw - 9uvw = 0【add 3uvw in each bracket then -9uvw at the end】
(u³ + v³ + w³) + 3uv (u + v + w) + 3vw (v + w + u) + 3uw (u + w + v) -3uvw = 0
(u³ + v³ + w³) + 3uv (0) + 3vw (0) + 3uw (0) -3uvw = 0【Since u+v+w=0】
u³ + v³ + w³ = 3uvw
Finally, taking cube again on both sides, we have the answer:
(u³ + v³ + w³)³ = 27u³ v³ w³


Hope it helps! ^^

2006-12-21 14:05:31 補充：
(2)So changing back into x, y, z,(x^(1/3)³＋y^(1/3)³＋z^(1/3)³)³＝27x^(1/3)³ y^(1/3)³ z^(1/3)³x＋y＋z＝27 x y z

2006-12-21 14:08:44 補充：
Sorry, some wrong typing in the supplement:(2)So changing back into x, y, z,(x^(1/3)³＋y^(1/3)³＋z^(1/3)³)³＝27x^(1/3)³ y^(1/3)³ z^(1/3)³(x＋y＋z)³＝27xyz