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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) (((2^3 −1)......(2019^3 −1))/((2^3 +1)......(2019^3 +1)))=Π_(k=2) ^(2019) (((k^3 −1))/((k^3 +1))) =Π_(k=2) ^(2019) (((k−1)(k^2 +k+1))/((k+1)(k^2 −k+1))) =Π_(k=2) ^(2019) ((k−1)/(k+1)).((Π_(k=2) ^(2019) (k^2 +k+1))/(Π_(j=1) ^(2018) ((j+1)^2 −(j+1)+1))) =((Π_(l=0) ^(2017) (l+2−1))/(Π_(k=2) ^(2019) (k+1))).((Π_(k=2) ^(2019) (k^2 +k+1))/(Π_(j=1) ^(2018) (j^2 +j+1))) =((Π_(l=0) ^(2017) (l+1))/(Π_(k=2) ^(2019) (k+1))).((Π_(k=2) ^(2019) (k^2 +k+1))/(Π_(j=1) ^(2018) (j^2 +j+1))) =((1.2)/((2020).(2019))).((2019^2 +2019+1)/3)

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

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

Beautiful

$$\:\mathrm{Beautiful} \\ $$

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

thank you

$${thank}\:{you} \\ $$

Commented by Algoritm last updated on 08/Nov/20

.

$$. \\ $$

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