Don't be confused!

$\Rightarrow x^4-x^2+1=0$

Multiplying by $x^2$ both sides.

$x^6-x^4+x^2=0$

$x^6=x^4-x^2$

We need to find: $x^5+\dfrac{1}{x}=\dfrac{x^6+1}{x}$

$\dfrac{x^4-x^2+1}{x}=\dfrac{0}{x}=\boxed{0}$

$x^4 - x^2 + 1 = 0 \quad \Rightarrow \begin{cases} x^4 = x^2 - 1 & \Rightarrow \color{#3D99F6}{x^5 = x^3 - x} \\ x^3 - x + \dfrac{1}{x} = 0 & \Rightarrow \color{#3D99F6}{\dfrac{1}{x} = -x^3 +x} \end{cases} \quad \Rightarrow x^5 + \dfrac{1}{x} = \boxed{0}$

I have a simpler one: :)

Given equation :

$x^4 - x^2 + 1 = 0$ --------- $\boxed{1}$

divide equation 1 by $x$

we get

$x^{3}-x+\frac{1}{x}=0$ , from here

$\frac{1}{x}=x-x^{3}$ ...... $\boxed{2}$

Multiply eq. 1 by $x$

we get:

$x^{5}-x^{3}+x=0$ , from here

$x^{5}=x^{3}-x$ ....... $\boxed{3}$

Add eq. 2 and 3

we get: $\frac{1}{x}+x^{5}=x-x^{3}+x^{3}-x$ = $\boxed{0}$

Nice solution by Abhay. Just highlighting a solution using complex numbers.
$x^{4} - x^{2} + 1 = 0$
$x^{2} + \dfrac{1}{x^{2}} = 1$
Let $x^{2} = e^{iy}$
$e^{iy} + e^{-iy} = 1$
$\cos(y) + i\sin(y) + \cos(y) -i\sin(y) = 1$
$\cos(y) = \dfrac{1}{2}$
Therefore, $x= e^{i\frac{\pi}{6}}$ is a root of this equation,

$x^{5} + \dfrac{1}{x} = e^{i\frac{5i\pi}{6}} + e^{\frac{-i\pi}{6}} = e^{i\pi}\cdot e^{\frac{-i\pi}{6}} + e^{\frac{-i\pi}{6}} = e^{\frac{-i\pi}{6}}(e^{i\pi}+1) = e^{\frac{-i\pi}{6}} \cdot 0 = 0$

Notice that:

$\left( x + \dfrac{1}{x} \right) \left(x^4-x^2+1 \right) = x^5 + x^3 - x^3 - x + x + \dfrac{1}{x} = x^5 + \dfrac{1}{x}$

Since we know that $x^4 - x^2 + 1 = 0$ , we can substitute it in:

$\left( x + \dfrac{1}{x} \right) \left(0 \right) = x^5 + \dfrac{1}{x}$

$\implies x^5 + \dfrac{1}{x} = \boxed{0}$

We observe that $x=0$ is not a root of the given equation. Thus, multiplying the given equation by $\displaystyle \frac{(x^{2}+1)}{x}$ and using the factorisation formula for $a^{3}+b^{3}$ : $\displaystyle \frac{(x^{2}+1)(x^{4}-x^{2}+1)}{x} = 0 \Rightarrow \boxed{x^5 + \frac{1}{x} = 0}$

${ \quad x }^{ 4 }+1=\quad { x }^{ 2\quad }\quad \\ \therefore { \quad x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =\quad 1\quad \\ \therefore \quad { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2=3\\ \therefore \quad \left( x+\frac { 1 }{ x } \right) =\pm \sqrt { 3 } \\ \therefore \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) \left( x+\frac { 1 }{ x } \right) =\quad \pm \sqrt { 3 } \\ { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =\quad 0\\ \therefore \quad { x }^{ 5 }+\frac { 1 }{ x } =\quad { x }^{ 2 }\left( { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } \right) =\quad \boxed { 0 } \\ \\ \\ \\$

Take x^2 common and then show that x^3+1/x^3 =0and the question is solved

Clearly the roots of the above equation are x^2 = -w , -w^(2) which are cube roots of unity . Put any of them above to get zero directly