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Question-164924




Question Number 164924 by cortano1 last updated on 23/Jan/22
Commented by bobhans last updated on 23/Jan/22
 (1) m^3 =2020+(√(2019))          n^3  = 2017 +(√(2019))         m^3 −n^3  = 3 ; 3+n^3 −m^3  = 0   (2) 3^3 +(n^3 )^3 +(−m^3 )^3  = 3(−3m^3 n^3 )           27+n^9 −m^9 = −9m^3 n^3            27 = m^9 −9m^3 n^3 −n^9
$$\:\left(\mathrm{1}\right)\:\mathrm{m}^{\mathrm{3}} =\mathrm{2020}+\sqrt{\mathrm{2019}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{n}^{\mathrm{3}} \:=\:\mathrm{2017}\:+\sqrt{\mathrm{2019}} \\ $$$$\:\:\:\:\:\:\:\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} \:=\:\mathrm{3}\:;\:\mathrm{3}+\mathrm{n}^{\mathrm{3}} −\mathrm{m}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{3}^{\mathrm{3}} +\left(\mathrm{n}^{\mathrm{3}} \right)^{\mathrm{3}} +\left(−\mathrm{m}^{\mathrm{3}} \right)^{\mathrm{3}} \:=\:\mathrm{3}\left(−\mathrm{3m}^{\mathrm{3}} \mathrm{n}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{27}+\mathrm{n}^{\mathrm{9}} −\mathrm{m}^{\mathrm{9}} =\:−\mathrm{9m}^{\mathrm{3}} \mathrm{n}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{27}\:=\:\mathrm{m}^{\mathrm{9}} −\mathrm{9m}^{\mathrm{3}} \mathrm{n}^{\mathrm{3}} −\mathrm{n}^{\mathrm{9}} \\ $$

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