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We need to show that there doesn't exist positive integer $k$ such that $k^3 + 2k^2 + 2k + 1$ is a cube number.

I tried solving it in this way: Firstly let the cube number we are looking for be $l$, we can write $l = k + d$, then it must hold $k^3 + 2k^2 + 2k+1=(k+d)^3$. If we extend the binom $(k+d)$ on third power $k^3+2k^2+2k+1 = k^3+3k^2d+3kd^2+d^3$. In order for this to be true it must be that $2 = 3d, 2=3d^2 \text{ and }d^3=1$. Surely this is contradiction, since such $d$ doesn't exist.

Is my work good enough or are there any mistakes that make it wrong, are there easier ways to show this?

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    $\begingroup$ With $k=-1$, we have $k^3+2k^2+2k+1=-1+2-2+1=0=0^3$. With $k=0$, we have $0^3+0+0+1=1=1^3$. $\endgroup$ – Hagen von Eitzen Oct 19 at 10:45
  • $\begingroup$ I just edited the question, I forgot to mention that $k$ must be positive. $\endgroup$ – someone123123 Oct 19 at 11:00
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Hint :

For all $K \in \mathbb N,$

$$k^3 + 2k^2 + 2k + 1 \gt(k)^3$$

And

$$k^3 + 2k^2 + 2k + 1 \lt k^3 + 3k^2 + 3k + 1 \implies k^3 + 2k^2 + 2k + 1 \lt (k+1)^3$$

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  • $\begingroup$ We only care about positive integer $k$ $\endgroup$ – someone123123 Oct 19 at 10:58
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    $\begingroup$ @someone123123 you should specify this in the question, $\endgroup$ – The Demonix _ Hermit Oct 19 at 10:59
  • $\begingroup$ I'm sorry about this mistake, I will correct it now. $\endgroup$ – someone123123 Oct 19 at 10:59
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Your argument is fallacious because an equation $$k^3+ak^2+bk+c=k^3+dk^2+ek+f$$ does not imply that coefficient-wise $a=d$, $b=e$, $c=f$. Such a conclusion would be valid only if the equation were known to hold for enough (three, in this case) distinct values of $k$.

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