+2651 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.
+16 \((a-b)^3=a^3-3a^2b+3ab^2-b^3\)=
\(1-3ab(a-b)\)=
\(1-3ab\)=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!
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