If ${\left(x+\frac{1}{x}\right)}^2=3$, then the value of $x^{2020}+x^{2021}+x^{2022}+x^{2014}+x^{2015}+x^{2016}+1$ is

Given:
$${\left(x+\frac{1}{x}\right)}^2=3⟹x+\frac{1}{x}=\sqrt{3}$$

Cubing both sides of the equation:

$${\left(x+\frac{1}{x}\right)}^3=(\sqrt{3}{)}^3$$

$$x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=3\sqrt{3}$$

Substitute $x+\frac{1}{x}=\sqrt{3}$:

$$x^3+\frac{1}{x^3}+3\sqrt{3}=3\sqrt{3}$$

$$x^3+\frac{1}{x^3}=0$$

Multiplying by $x^3$:

$$x^6+1=0⟹x^6=-1$$

Now rearrange the target expression by grouping terms with a power difference of $6$:

$$\text{Expression}=(x^{2022}+x^{2016})+(x^{2021}+x^{2015})+(x^{2020}+x^{2014})+1$$

$$=x^{2016}(x^6+1)+x^{2015}(x^6+1)+x^{2014}(x^6+1)+1$$

Since $x^6+1=0$, the first three terms vanish completely:

$$=x^{2016}(0)+x^{2015}(0)+x^{2014}(0)+1=1$$