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The real numbers a and b satisfy 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.

 Aug 17, 2024
 #1
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\((a-b)^3=a^3-3a^2b+3ab^2-b^3\)=

\(1-3ab(a-b)\)=

\(1-3ab\)=1

3ab=0,

ab=0

Because ab=0, we know \((a-b)^2\)=\(a^2+b^2+0\)=1

Because of this, \((a+b)^2=1+0=1\), so (a+b) is +/-1

Finally because a+b is 1, and one of either a or b is 0, because ab is 0, the possible pairings of (a,b) are

(0,1), or (1,0)

 

Hope this helps!

 ðŸ¤™ ðŸ¤™ ðŸ¤™

 Aug 17, 2024
edited by CelsiusSolving  Aug 17, 2024

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