Let $$\alpha$$ and $$\beta$$ be the roots of the equation, $$5x^2 + 6x - 2 = 0$$. If $$S_n = \alpha^n + \beta^n$$, $$n = 1, 2, 3, \ldots$$, then:

We begin with the quadratic equation whose roots are $$\alpha$$ and $$\beta$$:

$$5x^2 + 6x - 2 = 0.$$

Because $$\alpha$$ is a root, it satisfies the equation, so we write

$$5\alpha^2 + 6\alpha - 2 = 0.$$

Rearranging, we have

$$5\alpha^2 = -6\alpha + 2 \quad\Longrightarrow\quad \alpha^2 = \frac{-6\alpha + 2}{5}.$$

The same relation holds for $$\beta$$:

$$5\beta^2 + 6\beta - 2 = 0 \quad\Longrightarrow\quad \beta^2 = \frac{-6\beta + 2}{5}.$$

Our goal is to connect higher powers of $$\alpha$$ and $$\beta$$ to lower ones so that we can express the sum $$S_n = \alpha^n + \beta^n$$ recursively.

To do this, we multiply the root relation by higher powers. For $$\alpha$$, we multiply the equation $$5\alpha^2 + 6\alpha - 2 = 0$$ by $$\alpha^{\,n}$$:

$$5\alpha^{\,n+2} + 6\alpha^{\,n+1} - 2\alpha^{\,n} = 0.$$

A completely analogous step for $$\beta$$ gives

$$5\beta^{\,n+2} + 6\beta^{\,n+1} - 2\beta^{\,n} = 0.$$

Now we add the two equations term by term. On the left side, alike powers of $$\alpha$$ and $$\beta$$ combine naturally:

$$5\bigl(\alpha^{\,n+2} + \beta^{\,n+2}\bigr) + 6\bigl(\alpha^{\,n+1} + \beta^{\,n+1}\bigr) - 2\bigl(\alpha^{\,n} + \beta^{\,n}\bigr) = 0.$$

Using the definition $$S_k = \alpha^k + \beta^k$$, we rewrite each grouped term:

$$5S_{\,n+2} + 6S_{\,n+1} - 2S_{\,n} = 0.$$

This is the desired recurrence relation linking three consecutive sums of powers.

We now choose a specific value of $$n$$ that will involve $$S_6, S_5,$$ and $$S_4$$ because every option in the question features exactly these three. Setting $$n = 4$$ in the recurrence gives

$$5S_{\,4+2} + 6S_{\,4+1} - 2S_4 = 0 \quad\Longrightarrow\quad 5S_6 + 6S_5 - 2S_4 = 0.$$

Finally, we isolate the terms as they appear in the multiple-choice options:

$$5S_6 + 6S_5 = 2S_4.$$

This equality matches exactly Option C in the list.

Hence, the correct answer is Option C.