Five roots of $x^5+x+1=0$ and the value of $\prod_{k=1}^{5} (2+x_k^2)$ 
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*Here, $x_{k}$ are five roots of $x^{5} + x + 1 = 0$.

*I know two roots  are $\omega, \omega^{2}$ and next I can find a cubic dividing it by $x^{2} + x + 1$ and using the connection of $3$ roots with the coefficients of this cubic( Vieta's formulas ).

*But the calculation becomes very tedious, where I do not get the required value of $\prod_{k = 1}^{5}\left(2 + x_{k}^{2}\right) = 51$.

Can there be a simpler way of doing this ?.
 A: Let us transform $x^5+x+1=0$, by $y=2+x^2 \implies x=(y-2)^{1/2}$,
Then we get $$(y-2)^{5/2}+(y-2)^{1/2}=-1$$ sqyarinf this w have
$$(y-2)^5+(y-2)+2(y-2)^3-1=0$$
The required expression is nothing but the product of roots of this $y$-equation namely $y_1 y_2y_3y_4y_5$
Hence $$\prod_{k=1}^{5} (2+x_k^2)=y_1 y_2y_3y_4,y_5= -[-32-2 +2(-8)-1]= 51.$$
$$
A: Note that $\prod_{k=1}^5 (2+x_k^2) = \prod_{k=1}^5 (\sqrt{2}+ix_k)(\sqrt 2 - ix_k)$.
This is the product of all roots of a polynomial, whose roots are exactly $\sqrt{2} \pm ix_k$ for $k=1,...,5$.
Note that if $x^5+x+1$ has roots $x_1,...,x_5$, then $p(y) = (-iy+\sqrt 2i)^5 + (-iy+\sqrt 2i) + 1$ has roots $\sqrt 2 + ix_k$, $k=1,...,5$. The conjugate of this polynomial $\bar{p}$ has roots $\sqrt 2 - ix_k$.
Which means that the polynomial which has roots exactly equal to those we want, is $p\bar p$, and we need just the constant term of this whole polynomial, because by Vieta that is the product of all the roots. The constant of $p$ is $\sqrt 2^5i^5 + \sqrt 2i^5 +1 = 5\sqrt 2i + 1$, similarly of $\bar{p}$ is $1-5\sqrt 2 i$. Multiply these to get $1+(5\sqrt 2)^2 = 1+50=51$ and we are done.
A: Hint: $x^2+2=(x+\sqrt{2}i)(x-\sqrt{2}i)$
$x^5+x+1= (x-x_1)(x-x_2)\cdot...\cdot(x-x_5) \tag 1$
now put  $x=\sqrt{2}i$, $x=-\sqrt{2}i$ in $(1)$  and multiply both equations.
A: The question is pretty straight-forward. In questions, like these, sometimes, transformations work. For example, $\ { y }_{ i }={{ x }_{ i }}^{2}+2$ is a possibility but then, again, simplifying the expression would be tough. So, the below method is to be adopted.
$$\prod _{ k=1 }^{ 5 }{ ({ { x }_{ k } }^{ 2 }+2)= } \prod _{ k=1 }^{ 5 }{ ({ { x }_{ k } }+i\sqrt { 2 } )({ { x }_{ k } }-i\sqrt { 2 } )=(-f(-i\sqrt { 2 } ))*(-f(i\sqrt { 2 } )=f(i\sqrt { 2 } )f(-i\sqrt { 2 } )=(i*{ 2 }^{ \frac { 5 }{ 2 }  }+i*{ 2 }^{ \frac { 1 }{ 2 }  }+1)(-i*{ 2 }^{ \frac { 5 }{ 2 }  }-i*{ 2 }^{ \frac { 1 }{ 2 }  }+1)=1-{ (i*{ 2 }^{ \frac { 5 }{ 2 }  }+i*{ 2 }^{ \frac { 1 }{ 2 }  }) }^{ 2 }=1+{ (\sqrt { 2 } ) }^{ 2 }{ ({ 2 }^{ 2 }+1) }^{ 2 }=51 } $$
where $\ f(x)={ x }^{ 5 }+x+1=\prod _{ k=1 }^{ 5 }{ (x-{ x }_{ k }) } $
Hope it is helpful.
