fUNCTION

Assume there exist x1, x2 € R such that x1 is not equal to x2, and f(x1) = f(x2)

Since f(x1) = f(x2)

x1^3 = x2^3

x1^3 - x2^3 = 0

(x1 - x2)(x1^2 + x1x2 + x2^2) = 0

(x1 - x2)[(x1 - x2)^2 + x1x2] = 0

=> x1 = x2 [Since (x1 - x2)^2 + x1x2 =/= 0]

So, f is injective.

2009-02-18 20:25:32 補充：
For (x1 - x2)^2 + x1x2 =/= 0

Case I: If x1, x2 > 0, then it is obviously true.
Case II: If x1, x2 < 0, then it is also obviously true.

2009-02-18 20:25:35 補充：
Case III: Without loss of generosity, let x1 < 0 and x2 > 0
Then x1 - x2 = +-sqrt(x1x2)
But x1x2 is negative, in which the sqrt is not a real number, which is impossible.

Hence, the conclusion is (x1 - x2)^2 + x1x2 =/= 0

2009-02-18 20:32:40 補充：
In case III, since x1 - x2 is a real number, whereas sqrt(x1x2) is imaginary, so this case is impossible.