+1725 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 $b.$
+394 A)
\({(a-b)}^{3}= {a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3} = 1\)
\(1-3{a}^{2}b+3a{b}^{2}=1\)
\(3ab(b-a)=0\)
\(-3ab=0\)
\(ab=0\)
+394 B)
\({(a+b)}^2={(a-b)}^{2}+4ab\)
From previously, ab=0
\((a+b)^2={1}^{2}\)
\(a+b=\pm1\).
+394 C)
We know a-b = 1, and a+b = 1 or a+b = -1.
Consider two cases:
Case 1:
\(\begin{cases} a-b = 1\\ a+b=1 \end{cases}\)
\(2a = 2, a=1\)
\(\begin{cases} a=1 \\ b=0 \end{cases}\)
Case 2:
\(\begin{cases} a-b=1 \\ a+b = -1 \end{cases}\)
\(2a=0, a=0\)
\(\begin{cases} a=0 \\ b=-1 \end{cases}\)
Our two solutions are \(a, b = (1, 0)\) and \(a, b = (0, -1)\).
