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Question Number 90139 by jagoll last updated on 21/Apr/20

x = 2021^3 −2019^3 (√((x−2)/6)) = ?

$$\mathrm{x}\:=\:\mathrm{2021}^{\mathrm{3}} −\mathrm{2019}^{\mathrm{3}} \\ $$$$\sqrt{\frac{\mathrm{x}−\mathrm{2}}{\mathrm{6}}}\:=\:? \\ $$

Commented by jagoll last updated on 21/Apr/20

x = (p+2)^3 −p^3 , p = 2019 x = 6p^2 +12p+8 x−2 = 6p^2 +12p+6 x−2 = 6(p+1)^2 (√((x−2)/6)) = (√((6(p+1)^2 )/6)) = p+1 = 2019+1 = 2020

$$\mathrm{x}\:=\:\left(\mathrm{p}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{p}^{\mathrm{3}} \:,\:\mathrm{p}\:=\:\mathrm{2019} \\ $$$$\mathrm{x}\:=\:\mathrm{6p}^{\mathrm{2}} +\mathrm{12p}+\mathrm{8}\: \\ $$$$\mathrm{x}−\mathrm{2}\:=\:\mathrm{6p}^{\mathrm{2}} +\mathrm{12p}+\mathrm{6}\: \\ $$$$\mathrm{x}−\mathrm{2}\:=\:\mathrm{6}\left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{x}−\mathrm{2}}{\mathrm{6}}}\:=\:\sqrt{\frac{\mathrm{6}\left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$=\:\mathrm{p}+\mathrm{1}\:=\:\mathrm{2019}+\mathrm{1}\:=\:\mathrm{2020}\: \\ $$

Commented by MJS last updated on 21/Apr/20

good idea, just modify it a bit: t=2020 x=(t+1)^3 −(t−1)^3 =6t^2 +2 x−2=6t^2 (√((6t^2 )/6))=t=2020

$$\mathrm{good}\:\mathrm{idea},\:\mathrm{just}\:\mathrm{modify}\:\mathrm{it}\:\mathrm{a}\:\mathrm{bit}: \\ $$$${t}=\mathrm{2020} \\ $$$${x}=\left({t}+\mathrm{1}\right)^{\mathrm{3}} −\left({t}−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2} \\ $$$${x}−\mathrm{2}=\mathrm{6}{t}^{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{6}{t}^{\mathrm{2}} }{\mathrm{6}}}={t}=\mathrm{2020} \\ $$

Commented by jagoll last updated on 21/Apr/20

waw....very great sir. thank you

$$\mathrm{waw}....\mathrm{very}\:\mathrm{great}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$

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