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 ?.

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.
