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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

 Feb 6, 2023
 #1
avatar+2667 
+1

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 \(ab = \color{brown}\boxed{0}\)

 Feb 6, 2023
edited by BuilderBoi  Feb 6, 2023
 #2
avatar+2667 
+1

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 \(a + b = \color{brown}\boxed{-1, 1}\) 

 Feb 6, 2023
 #3
avatar+2667 
+1

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

 Feb 6, 2023

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