I return back

$\begin{aligned} x + \frac 1x & = 2 & \small \color{#3D99F6} \text{Multiply throughout by }x \text{ and rearrange.} \\ x^2 - 2x + 1 & = 0 \\ (x-1)^2 & = 0 \\ \implies x & = 1 \end{aligned}$

Therefore, $x^6 + \dfrac 1{x^6} = 1+1= \boxed{2}$

$\text{}x+$ $\frac{1}{x}=a$ $\implies$ $\text{}x^{6}+$ $\frac{1}{x^{6}}={a^{6}}-{6{a^{4}}}+{9{a^{2}}}-2$

we have :

$\text{}x+$ $\frac{1}{x}=2$ $\implies$ $\text{}x^{6}+$ $\frac{1}{x^{6}}=m$

Solve for $\text{}m$ .

For $\text{}x+$ $\frac{1}{x}=2$ $\implies$ $\text{}x^{6}+$ $\frac{1}{x^{6}}= {2^{6}}-{6×{2^{4}}}+{9×{2^{2}}}-2 ={\boxed{2}}$

x+1/x=2

(x+1/x)^2=4

x^2+1/x^2+2=4

x^2+1/x^2=2

(x^2+1/x^2)^3=8

x^6+1/x^6+3x1/x(x+1/x)=8

m+3*2=8

m=2

$\begin{aligned} x^6 + \dfrac 1{x^6} & = (x^3)^2 + \left(\dfrac 1{x^3}\right)^2 \\ & = \left(x^3 + \dfrac 1{x^3} \right)^2 -2 \\ & = \left\{\left(x + \dfrac 1x \right)^3 - 3\left(x + \dfrac 1x \right)\right \}^2 -2 \\ & = (2^3 - 3 \times 2)^2 - 2 \\ & = (8 - 6)^2 -2 \\ & = 2^2 -2 \\ & = 4 - 2 \\ & = \boxed 2.\\ \end{aligned}$