+1678 I think I know the answers, but I don't know how to prove it.
Let a and b be real numbers such that a - b = 1 and a^3 - b^3 = 1.
(a) Find all possible values of ab
(b) Find all possible values of a + b
(c) Find all possible values of a and b
+2668 a) Note that \(a^3 - b^3 = (a-b)(a^2 + b^2 + ab)\)
Substituting what we have, we get \(1 = 1 (a^2 + b^2 + ab)\), which means that \(a^2 + b^2 + ab = 1\)
We can write \(a^2 + b^2 + ab \) as \((a - b)^2 + 3ab\).
Using what we know, we have \(1^2 + 3ab = 1\), meaning \(ab = \color{brown}\boxed{0}\)
+2668 b) We already proved that \(a^2 + b^2 + ab = 1\).
Now, we write this as \((a+b)^2 - ab = 1\)
Because \(ab = 0\), we know \((a+b)^2 = 1\), meaning \(a + b = \color{brown}\boxed{-1, 1}\)
+2668 c) If ab = 0, either a or b = 0.
If a = 0, b = -1, so \((a, b)\) is \(\color{brown}\boxed{(0, -1)}\)
If b = 0, a = 1, so \(\color{brown}\boxed{(1, 0)}\)
These are the only solutions that work
