If $$a$$ and $$b$$ are the roots of the equation $$x^2 - 7x - 1 = 0$$, then the value of $$\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$$ is equal to _______

Let $$a$$ and $$b$$ be the two roots of $$x^{2}-7x-1=0$$.
Define the symmetric power sum $$S_{n}=a^{n}+b^{n}$$ for any positive integer $$n$$.

Case 1: Deriving a recurrence for $$S_{n}$$
Because $$a$$ and $$b$$ satisfy $$x^{2}-7x-1=0$$, we have $$a^{2}=7a+1$$ and $$b^{2}=7b+1$$.
Multiply these identities by $$a^{\,n-2}$$ and $$b^{\,n-2}$$ respectively (for $$n\ge 2$$):
$$a^{n}=7a^{\,n-1}+a^{\,n-2},\quad b^{n}=7b^{\,n-1}+b^{\,n-2}$$
Adding the two equalities gives the recurrence relation

$$S_{n}=7S_{n-1}+S_{n-2}\qquad\text{for }n\ge 2\;$$ $$-(1)$$

Case 2: Expressing $$S_{21}$$ in terms of lower sums
Use the recurrence twice:

$$S_{21}=7S_{20}+S_{19}\qquad\;$$ (from $$(1)$$ with $$n=21$$)
$$S_{20}=7S_{19}+S_{18}\qquad\;$$ (from $$(1)$$ with $$n=20$$)

Substitute the second equation into the first:

$$S_{21}=7\bigl(7S_{19}+S_{18}\bigr)+S_{19}=49S_{19}+7S_{18}+S_{19}=50S_{19}+7S_{18}$$ $$-(2)$$

Case 3: Expressing $$S_{17}$$ in terms of $$S_{19}$$ and $$S_{18}$$
Apply $$(1)$$ with $$n=19$$:

$$S_{19}=7S_{18}+S_{17}\;\;\Longrightarrow\;\;S_{17}=S_{19}-7S_{18}$$ $$-(3)$$

Case 4: Forming the required numerator
Add $$(2)$$ and $$(3)$$:

$$S_{21}+S_{17}=(50S_{19}+7S_{18})+(S_{19}-7S_{18})=51S_{19}$$ $$-(4)$$

Case 5: Computing the desired expression
The denominator is simply $$S_{19}$$, hence

$$\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}} =\frac{S_{21}+S_{17}}{S_{19}} =\frac{51S_{19}}{S_{19}} =51$$

Therefore, the required value is $$\mathbf{51}$$.