Let a and b with a > b > 0 be real numbers satisfying a^3 + b^3 = a + b. Find $\frac{(a + b)^2 - 1}{ab}$ .
+14915 Find \(\frac{(a + b)^2 - 1}{ab} \)
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\(a^3 + b^3 = a + b\\ a^3-a=b-b^3\)
\(\{a,b\}=\{-1,-1\},\{0,0\},\{1,1\}\)
\({\color{blue}\frac{(a + b)^2 - 1}{ab}}=\frac{(-1 + (-1))^2 - 1}{(-1)\cdot (-1)}\color{blue}=3 \\ {\color{blue}\frac{(a + b)^2 - 1}{ab}}=\frac{(1 + 1)^2 - 1}{1\cdot 1}\color{blue} =3\)
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