How find this sum $\sum_{k=1}^{n}\sqrt[3]{a_{k}}$ let sequence $\{a_{n}\}$ such 

$$(6k^2+2-a_{k})^3=27(4k^6+3k^4+3k^2-1)a_{k}$$
  for all $k\ge 1$, then compute
  $$\sum_{k=1}^{n}\sqrt[3]{a_{k}}$$

My try: I have note

$$(4k^6+3k^4+3k^2-1)=(2k-1)(2k+1)(k^2-k+1)(k^2+k+1)$$
  then
  $$6k^2+2-a_{k}=3\sqrt[3]{(2k-1)(2k+1)(k^2-k+1)(k^2+k+1)a_{k}}$$

Then I can't.Thank you very much
 A: An interesting observation is if one define $u_k=k^3+(k+1)^3$, then the defining equation of $a_k$ can be simplified as 
$$(u_k - u_{k-1} - a_k)^3 = 27 u_k u_{k-1} a_k\tag{*1}$$
Compare this with the algebraic identity
$$\left(x^3 - y^3 - (x-y)^3\right)^3 = 27 x^3 y^3 (x-y)^3$$
One will suspect
$$\sqrt[3]{a_k} = \sqrt[3]{u_k} - \sqrt[3]{u_{k-1}}$$
is a solution of $(*1)$. For $k \ge 1$ where $u_k > u_{k-1} > 0$, we can 
verify this is indeed the case by direct substitution. As a result,
$$\sum_{k=1}^n\sqrt[3]{a_k} = \sqrt[3]{u_n} - \sqrt[3]{u_0} = \sqrt[3]{n^3 + (n+1)^3} - 1$$
A: This looks rather intractable. It may help to know that:
$$a_k = 2 + 6 k^2 \\ +
  3 \left(1 - 3 k + 7 k^3 - 12 k^4 + 15 k^5 - 13 k^6 + 18 k^7 - 12 k^8 + 8 k^9\right)^{1/3} \\+ 
  3 \left(1 + 3 k - 7 k^3 - 12 k^4 - 15 k^5 - 13 k^6 - 18 k^7 - 12 k^8 - 8 k^9\right)^{1/3}$$
And so, developing $\sqrt[3]{a_k}$ into a power series, you can show that for large enough $k$:
$$\sqrt[3]{a_k} \approx 2^{1/3} - 2^{5/3} k^{-2}$$
So for large enough $n$, your sum should be pretty close to:
$$\sum_{k=1}^{n}\sqrt[3]{a_k}\approx2^{-5/3}(4 n-H_{n,2})$$
Of course, you could compute higher orders of the series if you really need the accuracy.
