Question Number 211043 by MATHEMATICSAM last updated on 26/Aug/24
Answered by Ghisom last updated on 26/Aug/24
$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{9}= \\ $$$$=\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{5}=\mathrm{2024} \\ $$
Answered by mr W last updated on 26/Aug/24
$${t}={x}−\mathrm{1}=\sqrt[{\mathrm{3}}]{\mathrm{2019}} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{9} \\ $$$$=\left({x}−\mathrm{1}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{6}\left({x}−\mathrm{1}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{12}\left({x}−\mathrm{1}+\mathrm{1}\right)+\mathrm{9} \\ $$$$=\left({t}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{6}\left({t}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{12}\left({t}+\mathrm{1}\right)+\mathrm{9} \\ $$$$={t}^{\mathrm{3}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{12}{t}+\mathrm{8}−\mathrm{6}{t}^{\mathrm{2}} −\mathrm{24}{t}−\mathrm{24}+\mathrm{12}{t}+\mathrm{12}+\mathrm{9} \\ $$$$={t}^{\mathrm{3}} +\mathrm{5} \\ $$$$=\mathrm{2019}+\mathrm{5} \\ $$$$=\mathrm{2024} \\ $$