Prove $\prod_{k=2}^N\left(1-\frac{2}{k^3+1}\right)~=~\frac23\left(1+\frac1{N(N+1)}\right)$

Question

I am struggling to verify the following

$$\prod_{k=2}^N\left(1-\frac{2}{k^3+1}\right)~=~\prod_{k=2}^N\left(\frac{k^3-1}{k^3+1}\right)~=~\frac23\left(1+\frac1{N(N+1)}\right)$$

I tried to use induction but failed at simplifing. My attempt so far

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First of all set $N=2$

$$\begin{align} \frac{2^3-1}{2^3+1}~&=~\frac23\left(1+\frac1{2(2+1)}\right)\\ \frac79~&=~\frac79 \end{align}$$

Next set $N=M$ and further $N=M+1$ to get

$$\begin{align} \prod_{k=2}^M\left(\frac{k^3-1}{k^3+1}\right)~&=~\frac23\left(1+\frac1{M(M+1)}\right)\\ \prod_{k=2}^{M+1}\left(\frac{k^3-1}{k^3+1}\right)~&=~\frac23\left(1+\frac1{(M+1)(M+2)}\right) \end{align}$$

By splitting up the second product and substitute the first formula in we get

$$\begin{align} \frac23\left(1+\frac1{(M+1)(M+2)}\right)~&=~\prod_{k=2}^{M+1}\left(\frac{k^3-1}{k^3+1}\right)\\ &=~\prod_{k=2}^M\left(\frac{k^3-1}{k^3+1}\right)\left(\frac{(M+1)^3-1}{(M+1)^3+1}\right)\\ &=~\frac23\left(1+\frac1{M(M+1)}\right)\left(\frac{(M+1)^3-1}{(M+1)^3+1}\right)\\ &=~\frac23\left(\frac{M(M+1)+1}{M(M+1)}\right)\left(\frac{(M+1)^3-1}{(M+1)^3+1}\right)\\ &=~\frac23\left(\frac{(M+1)^2-M}{(M+1)^2-(M+1)}\right)\left(\frac{(M+1)^3-1}{(M+1)^3+1}\right) \end{align}$$

And this were I got stuck. Removing the parentheses and then rewrite everthing leads nowhere. Could someone please help me with this task?

Thanks in advance.

Answer 1

$$1-\frac{2}{k^2+k+1}=\frac{k^3-1}{k^3+1} =\frac{k-1}{k+1}\frac{k^2+k+1}{k^2-k+1} =\frac{k-1}{k+1}\frac{k^2+k+1}{(k-1)^2+(k-1)+1}.$$ Then $$\prod_{k=2}^N\left(1-\frac{2}{k^2+k+1}\right) =\prod_{k=2}^N\frac{k-1}{k+1}\times \prod_{k=2}^N\frac{k^2+k+1}{(k-1)^2+(k-1)+1}.$$ Both these products telescope: $$\prod_{k=2}^N\frac{k-1}{k+1}=\frac2{N(N+1)}$$ and $$\prod_{k=2}^N\frac{k^2+k+1}{(k-1)^2+(k-1)+1} =\frac{N^2+N+1}{3}$$ etc.