If $$\displaystyle\sum_{k=1}^{10} \dfrac{k}{k^4 + k^2 + 1} = \dfrac{m}{n}$$, where $$m$$ and $$n$$ are co-prime, then $$m + n$$ is equal to ______.

We need to evaluate $$\displaystyle\sum_{k=1}^{10} \dfrac{k}{k^4 + k^2 + 1}$$ and express it as $$\dfrac{m}{n}$$ where $$\gcd(m, n) = 1$$.

$$k^4 + k^2 + 1 = (k^2 + k + 1)(k^2 - k + 1)$$

Verification: $$(k^2 + k + 1)(k^2 - k + 1) = k^4 - k^2 + k^2 + k^2 - k + k - 1 + 1 = k^4 + k^2 + 1$$. Correct.

Since $$(k^2 + k + 1) - (k^2 - k + 1) = 2k$$:

$$\dfrac{k}{(k^2+k+1)(k^2-k+1)} = \dfrac{1}{2}\left(\dfrac{1}{k^2-k+1} - \dfrac{1}{k^2+k+1}\right)$$

Let $$a_k = \dfrac{1}{k^2 - k + 1}$$. Then $$k^2 + k + 1 = (k+1)^2 - (k+1) + 1$$, so $$\dfrac{1}{k^2+k+1} = a_{k+1}$$.

$$\sum_{k=1}^{10} \dfrac{k}{k^4+k^2+1} = \dfrac{1}{2}\sum_{k=1}^{10}(a_k - a_{k+1}) = \dfrac{1}{2}(a_1 - a_{11})$$

$$a_1 = \dfrac{1}{1^2 - 1 + 1} = \dfrac{1}{1} = 1$$

$$a_{11} = \dfrac{1}{11^2 - 11 + 1} = \dfrac{1}{121 - 11 + 1} = \dfrac{1}{111}$$

$$\dfrac{1}{2}\left(1 - \dfrac{1}{111}\right) = \dfrac{1}{2} \cdot \dfrac{110}{111} = \dfrac{55}{111}$$

Since $$\gcd(55, 111) = \gcd(55, 1) = 1$$ (using $$111 = 2 \times 55 + 1$$), we have $$m = 55$$ and $$n = 111$$.

$$m + n = 55 + 111 = 166$$

The correct answer is $$\boxed{166}$$.