Polynomial saga

It is given that $x^4 + \dfrac{1}{x^4} = 194$ .

$\left( x^2 + \dfrac{1}{x^2} \right) ^2 = x^4 + 2 + \dfrac{1}{x^4} = 196$

$\Rightarrow x^2 + \dfrac{1}{x^2} = \sqrt{196} = 14$

Similarly, $x + \dfrac{1}{x} = \sqrt{14+2} = 4$

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

$\Rightarrow x^3 + \dfrac{1}{x^3} = \left( x^2 + \dfrac{1}{x^2} \right) \left( x + \dfrac{1}{x} \right) - \left( x + \dfrac{1}{x} \right) = 14 \times 4 - 4 = 52$

$x^6 + \dfrac{1}{x^6} = \left( x^3 + \dfrac{1}{x^3} \right) ^2 -2 = 52^2 - 2 = 2702$

Similarly, $x^{12} + \dfrac{1}{x^{12}} = 2702^2 -2 = 7300802$

Therefore,

$x^{12} + x^4 + x^3 + x^2 + x + \dfrac{1}{x^{12}} + \dfrac{1}{x^4} + \dfrac{1}{x^3} + \dfrac{1}{x^2} + \dfrac{1}{x}$

$= 7300802 + 194 + 52 + 14 + 4 = \boxed{7301066}$

Given that $\boxed{x^4+\dfrac{1}{x^4}=194}.$

$\Rightarrow (x^2+\dfrac{1}{x^2}) ^2 = 14^2 \Rightarrow \boxed{x^2+\dfrac{1}{x^2}=14}.$

Now , from the conclusion that $x^2+\dfrac{1}{x^2}=14,$ we will add $2$ on both sides .

Now , $(x+\dfrac{1}{x})^2=4^2 \Rightarrow\boxed { x+\dfrac{1}{x} = 4}.$

$x^3+\dfrac{1}{x^3} + 3(x+\dfrac{1}{x})=(x+\dfrac{1}{x})^3 \Rightarrow \boxed{x^3+\dfrac{1}{x^3}=52}.$

Squaring $x^4+\dfrac{1}{x^4}=194$ on both sides ,We get,

$x^8+\dfrac{1}{x^8}=37636-2 \Rightarrow x^8+\dfrac{1}{x^8}=37634.$

Now, $(x^8+\dfrac{1}{x^8})(x^4+\dfrac{1}{x^4})=x^{12}+\dfrac{1}{x^{12}}+x^4+\dfrac{1}{x^4}.$

Now, by substituting the respective values ,we get,

$\boxed{x^{12}+\dfrac{1}{x^{12}}=7300802}.$

Now,the given question is find ${ x }^{ 12 }+{ x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+{ \dfrac { 1 }{ { x }^{ 12 } } +\dfrac { 1 }{ { x }^{ 4 } } }+{ \dfrac { 1 }{ { x }^{ 3 } } }+{ \dfrac { 1 }{ { x }^{ 2 } } }+{ \dfrac { 1 }{ { x } } }$ ,

Which can easily be obtained by adding all the values in the boxes,that is $7300802+194+52+14+4 = 7301066.$