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Question Number 196650 by CrispyXYZ last updated on 28/Aug/23 $$\mathrm{Prove}\:\mathrm{that}\:{x}^{\mathrm{3}} ={y}^{\mathrm{6}} +\mathrm{20}\:\mathrm{has}\:\mathrm{no}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{solution}. \\ $$ Answered by AST last updated on 28/Aug/23 $${x}^{\mathrm{3}} \equiv\mathrm{1},−\mathrm{1}\:{or}\:\mathrm{0}\left({mod}\:\mathrm{9}\right) \\ $$$${when}\:\left({y},\mathrm{9}\right)=\mathrm{1};{y}^{\mathrm{6}}…

Question Number 196539 by qaz last updated on 27/Aug/23 $$\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{f}\left({i}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{f}\left(\mathrm{6}+\mathrm{1}−{i}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}{f}\left({i},{j}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}{f}\left(\mathrm{6}+\mathrm{1}−{j},\mathrm{6}+\mathrm{1}−{i}\right) \\…

Question Number 196594 by sonukgindia last updated on 27/Aug/23 Answered by JDamian last updated on 27/Aug/23 $${really}? \\ $$ Answered by sniper237 last updated on…

Question Number 196565 by mokys last updated on 27/Aug/23 $${find}\:{the}\:{power}\:{series}\:{exponition}\:{of}\: \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}\:\:{about}\:{z}_{{o}} \:=\:{i} \\ $$ Answered by aleks041103 last updated on 27/Aug/23 $$\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}−\mathrm{2}\right)\left({z}−\mathrm{1}\right)}=\frac{{A}}{{z}−\mathrm{1}}+\frac{{B}}{{z}−\mathrm{2}}…

Question Number 196498 by Pedro2002 last updated on 26/Aug/23 Answered by Frix last updated on 26/Aug/23 $$\int\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\int\frac{\mathrm{2}}{\:\sqrt{\pi}}\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\:{x}\:+{C} \\ $$ Terms of…