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Question Number 203245 by hardmath last updated on 13/Jan/24

If x = ((2019))^(1/3) + 1 Find: (x + 1)^3 −6∙(x + 1)^2 + 12x − 3

$$\mathrm{If}\:\:\:\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2019}}\:+\:\mathrm{1} \\ $$$$\mathrm{Find}: \\ $$$$\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{12x}\:−\:\mathrm{3} \\ $$

Answered by mr W last updated on 13/Jan/24

say t=x−1=((2019))^(1/3) ⇒t^3 =2019 (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+8−6(t^2 +4t+4)+12t+12+9 =t^3 +5 =2019+5 =2024

$${say}\:{t}={x}−\mathrm{1}=\sqrt[{\mathrm{3}}]{\mathrm{2019}} \\ $$$$\Rightarrow{t}^{\mathrm{3}} =\mathrm{2019} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{9} \\ $$$$=\left({x}−\mathrm{1}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{6}\left({x}−\mathrm{1}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{12}\left({x}−\mathrm{1}+\mathrm{1}\right)+\mathrm{9} \\ $$$$=\left({t}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{6}\left({t}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{12}\left({t}+\mathrm{1}\right)+\mathrm{9} \\ $$$$={t}^{\mathrm{3}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{12}{t}+\mathrm{8}−\mathrm{6}\left({t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{4}\right)+\mathrm{12}{t}+\mathrm{12}+\mathrm{9} \\ $$$$={t}^{\mathrm{3}} +\mathrm{5} \\ $$$$=\mathrm{2019}+\mathrm{5} \\ $$$$=\mathrm{2024} \\ $$

Commented by hardmath last updated on 13/Jan/24

Dear professor, answer: 2012

$$\mathrm{Dear}\:\mathrm{professor},\:\mathrm{answer}:\:\mathrm{2012} \\ $$

Commented by mr W last updated on 13/Jan/24

can′t you see that i have changed the question purposely to +9 in order to get a nice 2024 instead of 2012?

$${can}'{t}\:{you}\:{see}\:{that}\:{i}\:{have}\:{changed}\:\:{the} \\ $$$${question}\:{purposely}\:{to}\:+\mathrm{9}\:{in}\:{order}\:{to} \\ $$$${get}\:{a}\:{nice}\:\mathrm{2024}\:{instead}\:{of}\:\:\mathrm{2012}? \\ $$

Answered by Rasheed.Sindhi last updated on 13/Jan/24

(x+1)^3 −6(x+1)^2 +12(x+1)−15 let x+1=y y^3 −6y^2 +12y−15 y^3 −6y(y−2)−15 y^3 −3y(y−2)−8−7 (y−2)^3 −7 (x+1−2)^3 −7 (((2019))^(1/3) + 1−1)^3 −7 2019−7=2012

$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{6}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{12}\left({x}+\mathrm{1}\right)−\mathrm{15} \\ $$$${let}\:{x}+\mathrm{1}={y} \\ $$$${y}^{\mathrm{3}} −\mathrm{6}{y}^{\mathrm{2}} +\mathrm{12}{y}−\mathrm{15} \\ $$$${y}^{\mathrm{3}} −\mathrm{6}{y}\left({y}−\mathrm{2}\right)−\mathrm{15} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}\left({y}−\mathrm{2}\right)−\mathrm{8}−\mathrm{7} \\ $$$$\left({y}−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{7} \\ $$$$\left({x}+\mathrm{1}−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{7} \\ $$$$\left(\sqrt[{\mathrm{3}}]{\mathrm{2019}}\:+\:\mathrm{1}−\mathrm{1}\right)^{\mathrm{3}} −\mathrm{7} \\ $$$$\mathrm{2019}−\mathrm{7}=\mathrm{2012} \\ $$

Commented by Rasheed.Sindhi last updated on 13/Jan/24

May be the book, in which the answer 2012 is written, was published in 2012 and by then 2012 was as nice as 2024 today. :)

$${May}\:{be}\:{the}\:{book},\:{in}\:{which}\:{the} \\ $$$${answer}\:\mathrm{2012}\:{is}\:{written},\:{was}\: \\ $$$${published}\:{in}\:\mathrm{2012}\:{and}\:{by}\:{then} \\ $$$$\:\mathrm{2012}\:{was}\:{as}\:{nice}\:{as}\:\mathrm{2024} \\ $$$$\left.{today}.\::\right) \\ $$

Commented by hardmath last updated on 13/Jan/24

My dear professors, I think years are not important)

$$\mathrm{My}\:\mathrm{dear}\:\mathrm{professors}, \\ $$$$ \\ $$I think years are not important)

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