Evaluate expression Let $a$  and $b$ be  positive  real  numbers such  that $$ a^4 + 3ab + b^4 = \frac{1}{ab}$$ Evaluate $ {(\frac{a}{b})}^{\frac{1}{3}}  + {(\frac{b}{a})}^{\frac{1}{3}} - {(2+\frac{a}{b})}^{\frac{1}{2}}$. I  tried   in   five  different   ways but  I  can't  to  solve. I   suppose   that   the   expression  it   is  not   constant...
 A: If $t = a/b$, you can solve the first equation explicitly for $b$ in terms of $t$: 
$$ b = {\frac { \left( {t}^{8/3}-{t}^{4/3}+1 \right) \sqrt {{t}^{4/3}+1}}{
\sqrt [6]{t} \left( {t}^{4}+1 \right) }}
$$
and the expression you want to evaluate is
$$ t^{1/3} + t^{-1/3} - (2 + t)^{1/2} $$
A: Let us take first $a=b=1/\sqrt 2$. Then the equation $F(a,b)=1$ is satisfied for
$$
F(a,b) = ab(a^4+3ab+b^4)\ ,
$$
and for this special choice the expression we need to evaluate, using the function
$$
G(a,b) = 
\left(\frac ab\right)^{\frac 13}  
+
\left(\frac ba\right)^{\frac 13}  
-
\left(2+\frac ab\right)^{\frac 12}\ ,
$$
is $G(1/\sqrt 2, 1/\sqrt 2) = 1+1-\sqrt 3$.
The implicit equation $F(a,b)=1$ has many other solutions with $a\ne b$, 
if $F(a,b)=1$, then $F(b,a)=1$, too,
and we obtain obviously different values $G(a,b)\ne G(b,a)$.
The problem suggests however that $G(a,b)$ is constant on $F(a,b)=1$, this is not the case.
Note:
Here is a parametrization that works to have $F=1$, and the values for $G$ can be obtained explicitly:
$$
F\left(\ 
\sqrt{\frac{t^5}{t^4 + 1}} 
\ ,\ 
\sqrt{\frac 1{(t^4 + 1) t}}
\ \right)
$$
$$
=
{\left(\frac{t^{10}}{{\left(t^{4} + 1\right)}^{2}} + 3 \, \sqrt{\frac{t^{5}}{t^{4} + 1}} \sqrt{\frac{1}{{\left(t^{4} + 1\right)} t}} + \frac{1}{{\left(t^{4} + 1\right)}^{2} t^{2}}\right)} \sqrt{\frac{t^{5}}{t^{4} + 1}} \sqrt{\frac{1}{{\left(t^{4} + 1\right)} t}}
$$
$$
=1\ .
$$
