Let $a$ and $b$ with $a>b>0$ be real numbers satisfying $a^3+b^3=a+b$. Find $\sqrt{\dfrac{a}{b}+\dfrac{b}{a}-\dfrac{1}{ab}}$.
a^3 +b^3 = (a + b) ( a^2 - ab + b^2)
So
(a + b) = (a + b) ( a^2 - ab + b^2)
1 = a^2 - ab + b^2
ab = a^2 + b^2 - 1
Simpliying the expression under the radical we get
(a^2 + b^2 - 1 ) / ab =
ab / ab =
1
And the sqrt of 1 = 1
