Prove (a²+b²+6)/(ab) is a perfect cube

[anemone]

The numbers $a,\,b$ and $\dfrac{a^2+b^2+6}{ab}$ are positive integers. Prove that $\dfrac{a^2+b^2+6}{ab}$ is a perfect cube.

 

[Albert1]

The numbers $a,\,b$ and $\dfrac{a^2+b^2+6}{ab}$ are positive integers. Prove that $\dfrac{a^2+b^2+6}{ab}$ is a perfect cube.

Spoiler

let k=$\dfrac{a^2+b^2+6}{ab}---(1)$
then: $2+\dfrac{6}{ab}\leq k\leq \dfrac {a^2+b^2+6ab}{ab}=\dfrac {b}{a}+\dfrac {a}{b}+6$
for each $a,b\in N , max (2+\dfrac {6}{ab})=8,(a=b=1)$
and $min(\dfrac {b}{a}+\dfrac{a}{b}+6)=8 (for \,\, each \,\, a=b>0)$
if $a,b,k \in N, \,\,then\,\, k=8$
$\therefore k=8 $ is a perfect cube

 

[anemone]

Spoiler

let k=$\dfrac{a^2+b^2+6}{ab}---(1)$
then: $2+\dfrac{6}{ab}\leq k\leq \dfrac {a^2+b^2+6ab}{ab}=\dfrac {b}{a}+\dfrac {a}{b}+6$
for each $a,b\in N , max (2+\dfrac {6}{ab})=8,(a=b=1)$
and $min(\dfrac {b}{a}+\dfrac{a}{b}+6)=8 (for \,\, each \,\, a=b>0)$
if $a,b,k \in N, \,\,then\,\, k=8$
$\therefore k=8 $ is a perfect cube

Very well done, Albert!(Yes)