<s>[INST] 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
 [/INST] 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.
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