+0  
 
0
89
1
avatar

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
 #1
avatar+13577 
0

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

 

Hello Guest!

 

\(a^3 + b^3 = a + b\\ a^3-a=b-b^3\)

 

\(\{a,b\}=\{-1,-1\},\{0,0\},\{1,1\}\)

 

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

laugh  !

 Jan 14, 2022
edited by asinus  Jan 14, 2022

22 Online Users

avatar