Question-218775

Question Number 218775 by sonukgindia last updated on 15/Apr/25

Answered by SdC355 last updated on 15/Apr/25

(√(x+22))∈Z 1^(1/3) , 8^(1/3) , 27^(1/3) , 64^(1/3) ...etc x=−21 , −14 , 5 , .... −(21)^3 −6(−21)^2 −12∙21−32≠1 −(14)^3 −6(−14)^2 −14∙12−32≠2 5^3 −6∙5^2 +12∙5−32=3 ∴ x=5

$$\sqrt{{x}+\mathrm{22}}\in\mathbb{Z} \\ $$$$\mathrm{1}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{8}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{27}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{64}^{\mathrm{1}/\mathrm{3}} …\mathrm{etc} \\ $$$${x}=−\mathrm{21}\:,\:−\mathrm{14}\:,\:\mathrm{5}\:,\:…. \\ $$$$−\left(\mathrm{21}\right)^{\mathrm{3}} −\mathrm{6}\left(−\mathrm{21}\right)^{\mathrm{2}} −\mathrm{12}\centerdot\mathrm{21}−\mathrm{32}\neq\mathrm{1} \\ $$$$−\left(\mathrm{14}\right)^{\mathrm{3}} −\mathrm{6}\left(−\mathrm{14}\right)^{\mathrm{2}} −\mathrm{14}\centerdot\mathrm{12}−\mathrm{32}\neq\mathrm{2} \\ $$$$\mathrm{5}^{\mathrm{3}} −\mathrm{6}\centerdot\mathrm{5}^{\mathrm{2}} +\mathrm{12}\centerdot\mathrm{5}−\mathrm{32}=\mathrm{3} \\ $$$$\therefore\:{x}=\mathrm{5} \\ $$

Commented by SdC355 last updated on 15/Apr/25

other solution (z+22)=(z^3 −6z^2 +12z−32)^3 and Solve about z

$$\mathrm{other}\:\mathrm{solution} \\ $$$$\left({z}+\mathrm{22}\right)=\left({z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{12}{z}−\mathrm{32}\right)^{\mathrm{3}} \\ $$$$\mathrm{and}\:\mathrm{Solve}\:\mathrm{about}\:{z} \\ $$

Answered by Hanuda354 last updated on 15/Apr/25

Facebook Tweet Pin

Question-218775

Leave a Reply Cancel reply