Let A and B be unequal positive numbers, a 2-b 2 = a 3-b 3, and prove that 1

Proof: because A 2-B 2 = A 3-B 3

So (a-b) (a+b) = (a-b) (a 2+ab+b 2)

Because a and b are two unequal positive numbers.

a+b=a^2+ab+b^2=(a+b)^2-ab( 1)

Because (a+b) 2 >; 4ab

So ab < (a+b) 2/4

so-ab & gt； -(a+b)^2/4

So (a+b) 2-ab > (a+b) 2-(a+b) 2/4 = 3 (a+b) 2/4)

Therefore, a+b >; 3(a+b)^2/4

And (a+b) 2 = ab+a+b > from (1); a+b

Get a+b >; 1 or a+b

1 < a+b<4/3 because A 2-B 2 = A 3-B 3, (a-b) (a-b) = (a 2+ab+b 2), and because a ≠ b.

A+B = A 2+AB+B 2, that is, A+B = (A+B) 2-AB, so AB = (A+B) 2-(A+B), and because a≠b, AB.

So (a+b) 2-(a+b)

Namely 0

And because A+B = A 2+B 2+AB = (A+B) 2-AB, (A+B) 2 = A+B+AB, and A >；; 0，b & gt0,

So (a+b) 2 = a+b+ab >; A+b, that is, (a+b) 2 > A+b, so a+b >; 1, in summary: 1

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