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Question-211043




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

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