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Given-x-3-3x-2-2-6x-2-2-8-0-then-x-5-41x-2-2022-




Question Number 166954 by cortano1 last updated on 03/Mar/22
 Given x^3 −3x^2 (√2) +6x−2(√2)−8=0   then x^5 −41x^2 +2022 =?
$$\:\mathrm{Given}\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} \sqrt{\mathrm{2}}\:+\mathrm{6x}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{8}=\mathrm{0} \\ $$$$\:\mathrm{then}\:\mathrm{x}^{\mathrm{5}} −\mathrm{41x}^{\mathrm{2}} +\mathrm{2022}\:=? \\ $$
Answered by som(math1967) last updated on 03/Mar/22
x^3 −3.x^2 .(√2)+3.x.((√2))^2 −((√2))^3 =8  ⇒(x−(√2))^3 =2^3   ⇒x=2+(√2)  ∴ x^5 −41x^2 +2022  =(2+(√2))^5 −41(2+(√2))^2 +2022  =32+5×16×(√2)+10×8×2+10×4×2(√2)   +5×2×4+4(√2)−41(4+4(√2)+2)+2022  =32+160+40+80(√2)+80(√2)+4(√2)  −246−164(√2)+2022  =2022−14=2008
$${x}^{\mathrm{3}} −\mathrm{3}.{x}^{\mathrm{2}} .\sqrt{\mathrm{2}}+\mathrm{3}.{x}.\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{8} \\ $$$$\Rightarrow\left({x}−\sqrt{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{2}^{\mathrm{3}} \\ $$$$\Rightarrow{x}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$\therefore\:{x}^{\mathrm{5}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{2022} \\ $$$$=\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{5}} −\mathrm{41}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2022} \\ $$$$=\mathrm{32}+\mathrm{5}×\mathrm{16}×\sqrt{\mathrm{2}}+\mathrm{10}×\mathrm{8}×\mathrm{2}+\mathrm{10}×\mathrm{4}×\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\:+\mathrm{5}×\mathrm{2}×\mathrm{4}+\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{41}\left(\mathrm{4}+\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{2}\right)+\mathrm{2022} \\ $$$$=\mathrm{32}+\mathrm{160}+\mathrm{40}+\mathrm{80}\sqrt{\mathrm{2}}+\mathrm{80}\sqrt{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$−\mathrm{246}−\mathrm{164}\sqrt{\mathrm{2}}+\mathrm{2022} \\ $$$$=\mathrm{2022}−\mathrm{14}=\mathrm{2008} \\ $$

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