A smarter way, I suppose

Let $a$ and $b$ be the two roots, then $ab= 1$ and

$a^5 - 209a + 56 = 0, \ \ \ \ \ \ \ \ \ b^5 - 209b + 56 = 0$

Or

$a^5 = 209a - 56, \ \ \ \ \ \ \ \ \ b^5 = 209b - 56$

Product of this two equation gives

$(ab)^5 = 209^2 (ab) - 209 \cdot 56 \cdot (a + b) + 56^2$

Substitution of $ab = 1$ yields $a + b = \boxed{4}$

Let the two roots be $\alpha$ and $\dfrac {1}{\alpha}$ , and their sum be $a = \alpha +\dfrac {1}{\alpha}$ ; then the two factors are:

$(x-\alpha)(x-\dfrac {1}{\alpha}) = x^2 - (\alpha +\dfrac {1}{\alpha})x +1 = x^2 - ax + 1$

Then, let:

$x^5 - 209x + 56 = (x^2 - ax +1)(x^3 + bx^2 + cx + d)$

$\quad \quad x^5 + (b-a)x^4+(c-ab+1)x^3 + (d-ac+b)x^2 + (c-ad)x + d$

Equating the coefficients:

$\Rightarrow \begin {cases} d = 56 & \\ b-a = 0 & \Rightarrow b = a \\ c-ab+1 = 0 & \Rightarrow c = a^2 -1 \\ c - ad = - 209 & \Rightarrow a^2 - 1 - 56a = - 209 \quad \Rightarrow a^2 - 56a + 208 = 0 \end {cases}$

From $\quad a^2 - 56a + 208 = (a-4)(a-52)=0$

$\Rightarrow \begin {cases} a = 4 & \Rightarrow \alpha +\frac {1}{\alpha} = 4 & \Rightarrow \alpha^2 - 4\alpha + 1 = 0 & \Rightarrow x = 2 \pm \sqrt{3} \\ a = 52 & \Rightarrow \alpha +\frac {1}{\alpha} = 52 & \Rightarrow \alpha^2 - 52\alpha + 1 = 0 & \Rightarrow x = 26 \pm 15 \sqrt{3} \end {cases}$

It can be shown that $x = 26 \pm 15 \sqrt{3}$ are not roots of the equation. Therefore the sum of the two roots $a = \boxed {4}$

Let the five roots of the equation are $a,b,c,d$ and $e$ .

Let the two roots whose product is $1$ are $a$ and $b$ , $a \cdot b = 1$ .

By Vieta's Formula , we have

1. $abcde = -56$ Since $ab =1$ then $cde = -56$

2. $a+b+c+d+e = 0 \rightarrow c+d+e = -(a+b)$

3. $ab+ac+ad+ae+bc+bd+be+cd+ce+df = 0$ $1+(a+b)c+(a+b)d+(a+b)e+cd+ce+de = 0$ $1+(a+b)(c+d+e)+(cd+ce+de) = 0$ Since $c+d+e=-(a+b)$ so $1-(a+b)^2+(cd+ce+de) = 0$ $cd+ce+de=(a+b)^2 - 1$

4. $abcd+abce+abde+acde+bcde = -209$ $(ab)(cd+ce+de)+(a+b)(cde) = -209$

Since $cd+ce+de=(a+b)^2 - 1$ and $cde = -56$ and $ab=1$ so

$(a+b)^2 - 1 - 56(a+b) + 209 = 0$

$(a+b)^2 - 56(a+b) + 208 = 0$

Solve $(a+b)^2 - 56(a+b) + 208 = 0$ we have

$((a+b)-4)((a+b)-52) = 0$ $a+b = 4$ or $a+b = 52$ Hence $a+b = \boxed{4}$