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Algebra

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

Jan 13, 2022

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Find $$\frac{(a + b)^2 - 1}{ab}$$

Hello Guest!

$$a^3 + b^3 = a + b\\ a^3-a=b-b^3$$

$$\{a,b\}=\{-1,-1\},\{0,0\},\{1,1\}$$

$${\color{blue}\frac{(a + b)^2 - 1}{ab}}=\frac{(-1 + (-1))^2 - 1}{(-1)\cdot (-1)}\color{blue}=3 \\ {\color{blue}\frac{(a + b)^2 - 1}{ab}}=\frac{(1 + 1)^2 - 1}{1\cdot 1}\color{blue} =3$$

!

Jan 14, 2022
edited by asinus  Jan 14, 2022