Wanna bash it!

$\begin{aligned} x^5+x+1 & = 0 \\ \Rightarrow (x^2+x+1)(x^3-x^2+1) & = 0 \end{aligned}$

Now, let the roots of: $\begin{cases} x^2 + x + 1 = 0 & \text{be } x_1, x_2 & \Rightarrow \begin{cases} x_1 + x_2 & = -1 \\ x_1x_2 & = 1 \end{cases} \\ x^3 - x^2 + 1 = 0 & \text{be } x_3, x_4, x_5 & \Rightarrow \begin{cases} x_3 + x_4 + x_5 & = 1 \\ x_3x_4 + x_4x_5 + x_5x_3 & = 0 \\ x_3x_4x_5 & = -1 \end{cases} \end{cases}$

Now, we have:

$\begin{aligned} (x_1^2+2)(x_2^2+2) & = x_1^2x_2^2+2(x_1^2+x_2^2)+4 \\ & = (x_1x_2)^2+2\left((x_1+x_2)^2 - 2x_1x_2 \right) + 4 \\ & = 1 + 2(1-2) + 4 \\ & = 3 \end{aligned}$

$\begin{aligned} (x_3^2+2)(x_4^2+2)(x_5^2+2) & = x_3^2x_4^2x_5^2 + 2\left(x_3^2x_4^2 + x_4^2x_5^2 + x_5^2x_3^2\right) + 4\left(x_3^2+x_4^2+x_5^2\right) + 8 \\ & = \left(x_3x_4x_5\right)^2 + 2\left((x_3x_4 + x_4x_5 + x_5x_3)^2 - 2(x_3^2x_4x_5 + x_3x_4^2x_5 + x_3x_4x_5^2) \right) \\ & \quad \quad + 4\left((x_3+x_4+x_5)^2 - 2(x_3x_4 + x_4x_5 + x_5x_3) \right) + 8 \\ & = 1 + 2(0) - 4(x_3x_4x_5)(x_3+x_4+x_5) +4\left(1-2(0)\right) + 8 \\ & = 13 - 4(-1)(1) \\ & = 17 \end{aligned}$

Therefore,

$\begin{aligned} (x_1^2+2)(x_2^2+2)(x_3^2+2)(x_4^2+2)(x_5^2+2) & = 3 \times 17 = \boxed{51} \end{aligned}$

$A\quad simplest\quad solution\quad according\quad to\quad me...\\ if\quad f(x)\quad has\quad roots\quad { x }_{ i }\quad 1\le i\le 5\\ Then\quad g(x)\quad with\quad roots\quad { x }_{ i }^{ 2 }+2\quad =\quad f(\pm \sqrt { x-2 } )\\ \Rightarrow { (x-2) }^{ \frac { 5 }{ 2 } }+{ (x-2) }^{ \frac { 1 }{ 2 } }+1=0\\ Take\quad 1\quad to\quad RHS\quad and\quad square\\ { ({ x }^{ 2 }-4x+5) }^{ 2 }(x-2)=1\\ Now\quad simply\quad find\quad product\quad of\quad roots\\ (no\quad need\quad to\quad expand)\quad \frac { -(-51) }{ 1 } =51$

The given equation can be written as, ${x}^5+x+1=(x-{x}_{1})(x-{x}_2)....(x-{x}_5)$

The above equation is satisfied for all values of $x$ as the graph of this function is for all values of $x$ . Therefore, This equation will be satisfied for $x=\sqrt{2}i$ and $x=-\sqrt{2}i$

Take $x=\sqrt{2}i$

Therefore, ${\sqrt{2}}^{5}*{i}^{5}+\sqrt{2}i+1=(\sqrt{2}i-{x}_{1})......(\sqrt{2}i-{x}_5)=5\sqrt{2}i+1$

Take $x=-\sqrt{2}i$

Therefore, ${-\sqrt{2}}^{5}*{i}^{5}-\sqrt{2}i+1=(-\sqrt{2}i-{x}_{1})......(-\sqrt{2}i-{x}_{5})=-5\sqrt{2}i+1$

Multiplying both the equations we get, $-[( \sqrt{2}i-{x}_{1})( \sqrt{2}i+{x}_{1}).........(\sqrt{2}i-{x}_{5})( \sqrt{2}i+{x}_{5})=(5\sqrt{2}i+1)(-5\sqrt{2}i+1)$

Simplifying this equation we get that, $(2+{x}_{1}^{2}).......(2+{x}_{5}^{2})=1-(-25*2)=\boxed{51}$

Let's denote $S_1=x_1+x_2+x_3+x_4+x_5$ , $S_2=x_1x_2+x_1x_3+...$ , $S_3=x_1x_2x_3+x_1x_2x_4+...$ , $S_4=x_1x_2x_3x_4+...$ and $S_5=x_1x_2x_3x_4x_5$ . From Vieta's formulae we have $S_1=S_2=S_3=0$ , $S_4=1$ and $S_5=-1$ .

We want to evaluate: $\prod_{k=1}^5 (x_k^2+2)$ .

Considering i as imaginary unit:

$\prod_{k=1}^5 (x_k^2+2)=\prod_{k=1}^5 (x_k+i\sqrt{2})(x_k-i\sqrt{2})=\prod_{k=1}^5 (x_k+i\sqrt{2})\prod_{k=1}^5 (x_k-i\sqrt{2})$

Doing this product combinating terms, we have:

$\prod_{k=1}^5 (x_k+i\sqrt{2})=S_5+i\sqrt{2} S_4+(i\sqrt{2})^2 S_3+(i\sqrt{2})^3 S_2+(i\sqrt{2})^4 S_1+(i\sqrt{2})^5$

$\prod_{k=1}^5 (x_k+i\sqrt{2})=-1+i\sqrt{2}+(i\sqrt{2})^5=-1+5\sqrt{2}i$

In the same way:

$\prod_{k=1}^5 (x_k-i\sqrt{2})=-1-i\sqrt{2}+(-i\sqrt{2})^5=-1-5\sqrt{2}i$

Thus:

$\prod_{k=1}^5 (x_k^2+2) = (-1+5\sqrt{2}i)(-1-5\sqrt{2}i) = ((-1)^2+(5\sqrt{2})^2) = 51$ .