Step-by-step Solution:

1. Analyze the given information. We are given the quadratic equation ${x^2} - 6x - 2 = 0$. Let its roots be $\alpha$ and $\beta$, with the condition $\alpha > \beta$. A sequence ${a_n}$ is defined as ${a_n} = {\alpha ^n} - {\beta ^n}$ for $\,n \ge 1$. We need to find the value of the expression ${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$.

2. Use the properties of the roots. Since $\alpha$ and $\beta$ are the roots of the equation ${x^2} - 6x - 2 = 0$, they must satisfy the equation. Therefore: ${\alpha^2} - 6\alpha - 2 = 0 \quad \implies \quad {\alpha^2} = 6\alpha + 2$ ${\beta^2} - 6\beta - 2 = 0 \quad \implies \quad {\beta^2} = 6\beta + 2$

3. Establish a recurrence relation for the sequence $a_n$. From the relations in step 2, we can derive a general recurrence. Multiply the first equation by ${\alpha^{n-2}}$ (for $n \ge 2$): ${\alpha^{n-2}}({\alpha^2} - 6\alpha - 2) = 0 \implies {\alpha^n} - 6{\alpha^{n-1}} - 2{\alpha^{n-2}} = 0$ $\implies {\alpha^n} = 6{\alpha^{n-1}} + 2{\alpha^{n-2}}$ Similarly, for the root $\beta$: ${\beta^n} = 6{\beta^{n-1}} + 2{\beta^{n-2}}$ Now, let's use the definition of ${a_n}$: $a_n = {\alpha^n} - {\beta^n}$ $a_n = (6{\alpha^{n-1}} + 2{\alpha^{n-2}}) - (6{\beta^{n-1}} + 2{\beta^{n-2}})$ $a_n = 6({\alpha^{n-1}} - {\beta^{n-1}}) + 2({\alpha^{n-2}} - {\beta^{n-2}})$ $a_n = 6{a_{n-1}} + 2{a_{n-2}}$ This recurrence relation is valid for $n \ge 3$ (or $n \ge 2$ if we define $a_0 = 0$).

4. Apply the recurrence relation to the expression. The expression we need to evaluate is ${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$. From the recurrence relation $a_n = 6{a_{n-1}} + 2{a_{n-2}}$, we can rearrange it as: $a_n - 2{a_{n-2}} = 6{a_{n-1}}$ Let's substitute $n = 10$ into this rearranged relation: ${a_{10}} - 2{a_{10-2}} = 6{a_{10-1}}$ ${a_{10}} - 2{a_8} = 6{a_9}$

5. Substitute the result back into the original expression. Now we can substitute the value of the numerator we found in the previous step: ${{{a_{10}} - 2{a_8}} \over {2{a_9}}} = {{6{a_9}} \over {2{a_9}}}$

6. Simplify to find the final answer. Assuming ${a_9} \neq 0$, we can cancel the term ${a_9}$ from the numerator and denominator. ${{6{a_9}} \over {2{a_9}}} = {6 \over 2} = 3$ (We can verify that $a_9 \neq 0$. The roots are $x = 3 \pm \sqrt{11}$. So $\alpha = 3 + \sqrt{11} > 0$ and $\beta = 3 - \sqrt{11} < 0$. Then $a_9 = \alpha^9 - \beta^9 > 0 - (\text{negative}) > 0$).

Alternative Method

1. Substitute the definition of $a_n$ directly. ${{{a_{10}} - 2{a_8}} \over {2{a_9}}} = {{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)} \over {2(\alpha^9 - \beta^9)}}$

2. Rearrange the terms in the numerator. $= {{(\alpha^{10} - 2\alpha^8) - (\beta^{10} - 2\beta^8)} \over {2(\alpha^9 - \beta^9)}}$ $= {{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)} \over {2(\alpha^9 - \beta^9)}}$

3. Use the root property. From Step 2 of the first method, we have: ${\alpha^2} - 6\alpha - 2 = 0 \implies {\alpha^2} - 2 = 6\alpha$ ${\beta^2} - 6\beta - 2 = 0 \implies {\beta^2} - 2 = 6\beta$

4. Substitute these back into the expression. $= {{\alpha^8(6\alpha) - \beta^8(6\beta)} \over {2(\alpha^9 - \beta^9)}}$ $= {{6\alpha^9 - 6\beta^9} \over {2(\alpha^9 - \beta^9)}}$

5. Simplify. $= {{6(\alpha^9 - \beta^9)} \over {2(\alpha^9 - \beta^9)}} = {6 \over 2} = 3$

Both methods yield the same result.

Final Answer: The value of the expression is 3.