Question Number 90139 by jagoll last updated on 21/Apr/20
$$\mathrm{x}\:=\:\mathrm{2021}^{\mathrm{3}} −\mathrm{2019}^{\mathrm{3}} \\ $$$$\sqrt{\frac{\mathrm{x}−\mathrm{2}}{\mathrm{6}}}\:=\:? \\ $$
Commented by jagoll last updated on 21/Apr/20
$$\mathrm{x}\:=\:\left(\mathrm{p}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{p}^{\mathrm{3}} \:,\:\mathrm{p}\:=\:\mathrm{2019} \\ $$$$\mathrm{x}\:=\:\mathrm{6p}^{\mathrm{2}} +\mathrm{12p}+\mathrm{8}\: \\ $$$$\mathrm{x}−\mathrm{2}\:=\:\mathrm{6p}^{\mathrm{2}} +\mathrm{12p}+\mathrm{6}\: \\ $$$$\mathrm{x}−\mathrm{2}\:=\:\mathrm{6}\left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{x}−\mathrm{2}}{\mathrm{6}}}\:=\:\sqrt{\frac{\mathrm{6}\left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$=\:\mathrm{p}+\mathrm{1}\:=\:\mathrm{2019}+\mathrm{1}\:=\:\mathrm{2020}\: \\ $$
Commented by MJS last updated on 21/Apr/20
$$\mathrm{good}\:\mathrm{idea},\:\mathrm{just}\:\mathrm{modify}\:\mathrm{it}\:\mathrm{a}\:\mathrm{bit}: \\ $$$${t}=\mathrm{2020} \\ $$$${x}=\left({t}+\mathrm{1}\right)^{\mathrm{3}} −\left({t}−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2} \\ $$$${x}−\mathrm{2}=\mathrm{6}{t}^{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{6}{t}^{\mathrm{2}} }{\mathrm{6}}}={t}=\mathrm{2020} \\ $$
Commented by jagoll last updated on 21/Apr/20
$$\mathrm{waw}….\mathrm{very}\:\mathrm{great}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$