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Question Number 70838 by A8;15: last updated on 08/Oct/19

Answered by mind is power last updated on 08/Oct/19

$$X^3-1=(X-1)(X^2+X+1)$$$$X^3+1=(X+1)(X^2-X+1)$$$$\frac{(2^3-1)......({2019}^3-1)}{(2^3+1)......({2019}^3+1)}=\underset{k=2}{\prod ^{2019}}\frac{(k^3-1)}{(k^3+1)}$$$$=\underset{k=2}{\prod ^{2019}}\frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}$$$$=\underset{k=2}{\prod ^{2019}}\frac{k-1}{k+1}.\frac{\underset{k=2}{\prod ^{2019}}(k^2+k+1)}{\underset{j=1}{\prod ^{2018}}({(j+1)}^2-(j+1)+1)}$$$$=\frac{\underset{l=0}{\prod ^{2017}}(l+2-1)}{\underset{k=2}{\prod ^{2019}}(k+1)}.\frac{\underset{k=2}{\prod ^{2019}}(k^2+k+1)}{\underset{j=1}{\prod ^{2018}}(j^2+j+1)}$$$$=\frac{\underset{l=0}{\prod ^{2017}}(l+1)}{\underset{k=2}{\prod ^{2019}}(k+1)}.\frac{\underset{k=2}{\prod ^{2019}}(k^2+k+1)}{\underset{j=1}{\prod ^{2018}}(j^2+j+1)}$$$$=\frac{1.2}{\left(2020\right).\left(2019\right)}.\frac{{2019}^2+2019+1}{3}$$

Commented by Prithwish sen last updated on 08/Oct/19

$$\mathrm{Beautiful}$$

Commented by mind is power last updated on 08/Oct/19

$$thankyou$$

Commented by Algoritm last updated on 08/Nov/20

$$.$$

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