Prove by math induction

Question

$\forall n \geq 2$

$\frac{7}{9} \times \frac{26}{28} \times \ldots \times \frac{n^3 -1}{n^3 + 1} = \frac{2}{3} \times (1 + \frac{1}{n(n+1)})$

After basis step i went this far:

$ \frac{2n(n+1)+2}{3n(n+1)} \times \frac{(n-2)(n^2-n+1)}{n(n^2-3n+3)} $

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I am stuck here don't know how to clear this out, thanks alot.

Answer 1

${2\over3}\times(1+{1\over n(n+1)})\times{(n^3-1)\over(n^3+1)}$

$={2\over3}\times{(n^2+n+1)\over n(n+1)}\times{(n^3-1)\over(n^3+1)}$

$={2\over3}\times{(n^2+n+1)\over n(n+1)}\times{n((n+1)^2+(n+1)+1)\over(n+2)((n+1)^2-(n+1)+1)}$

$={2\over3}\times{(n^2+n+1)\over(n+1)}\times{(n+1)^2+(n+1)+1\over(n+2)(n^2+n+1)}$

$={2\over3}\times{1\over(n+1)}\times{(n+2)(n+1)+1\over(n+2)}$

$={2\over3}\times{(n+2)(n+1)+1\over(n+2)(n+1)}$

$={2\over3}\times(1+{1\over(n+1)(n+2)}) $

Answer 2

If $\prod_{r=2}^m\dfrac{r^3-1}{r^3+1}=\dfrac23\cdot\dfrac{m^2+m+1}{m(m+1)},$

$\prod_{r=2}^{m+1}\dfrac{r^3-1}{r^3+1}=\dfrac{(m+1)^3-1}{(m+1)^3+1}\cdot\prod_{r=2}^m\dfrac{r^3-1}{r^3+1}$

$=\dfrac23\cdot\dfrac{m[(m+1)^2+(m+1)+1]}{(m+2)\{(m+1)^2-(m+1)+1\}}\cdot\dfrac{m^2+m+1}{m(m+1)}$

$=\dfrac23\cdot\dfrac{(m+1)^2+(m+1)+1}{(m+1+1)(m+1)}$

as $(m+1)^2-(m+1)+1=m^2+m+1\ne0$ for real $m$