Roots Everywhere

Let $a$ be equal to $\sqrt [ 3 ]{ x }$

So accordingly, we need to find the value of $a+\frac { 1 }{ a }$

$\sqrt [ 3 ]{ 3(a-\frac { 1 }{ a } ) } =2\\ 3(a-\frac { 1 }{ a } )=8\\ a-\frac { 1 }{ a } =\frac { 8 }{ 3 } \\ { a }^{ 2 }-2+{ (\frac { 1 }{ a } ) }^{ 2 }=\frac { 64 }{ 9 } \\ { a }^{ 2 }+2+{ (\frac { 1 }{ a } ) }^{ 2 }=\frac { 64 }{ 9 } +4\\ { (a+\frac { 1 }{ a } ) }^{ 2 }=\frac { 100 }{ 9 } \\ a+\frac { 1 }{ a } =\frac { 10 }{ 3 } \\ \sqrt [ 3 ]{ x } +\frac { 1 }{ \sqrt [ 3 ]{ x } } =\frac { 10 }{ 3 }$

let's say:

$a\quad=\quad \sqrt [ 3 ]{ x }$

so:

$\sqrt [ 3 ]{ 3(a-\frac { 1 }{ a } ) } = 2$

after cubing and opening:

$a-\frac { 1 }{ a }=\frac { 8 }{ 3 }$

$3{ a }^{ 2 }-8a-3\quad=\quad 0$

after solving the sqrt we get:

${ a }_{ 1 }\quad=\quad 3$

${ a }_{ 2 }\quad=\quad -\frac { 1 }{ 3 }$

and we get 2 solutions:

$a+\frac { 1 }{ a } \quad=\quad \frac { 10 }{ 3 } \quad or\quad -\frac { 10 }{ 3 }$

and I didn't understand why not $-\frac { 10 }{ 3 }$

Take $\sqrt[3]{x}=m$ , then the given equation becomes:

$\sqrt[3]{3(m - \frac{1}{m})}=2$

Cubing both sides,

$\implies 3(m - \frac{1}{m})=8$

$\implies m - \frac{1}{m} = \frac{8}{3}$

Multiplying both sides of the equation by $3m$ ,

$\implies 3m^2 - 3 = 8m$

$\implies 3m^2 -8m -3=0$

$\implies \boxed{m=3, \frac{-1}{3}}$

But recall that in the question a condition is given that $x>0$ , which means:

$\implies \sqrt[3]{x} > \sqrt[3]{0} = 0$

$\implies m>0$ .

So, we have to eliminate $\frac{-1}{3}$ and accept $3$ as the sole value satisfying the condition.

Moving forward,

$\sqrt[3]{x}+\dfrac{1}{\sqrt[3]{x}} = 3 + \frac{1}{3} = \boxed{\frac{10}{3}}$