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Question Number 166954 by cortano1 last updated on 03/Mar/22

$$\mathrm{Given}{\mathrm{x}}^3-{3\mathrm{x}}^2\sqrt{2}+6\mathrm{x}-2\sqrt{2}-8=0$$$$\mathrm{then}{\mathrm{x}}^5-{41\mathrm{x}}^2+2022=?$$

Answered by som(math1967) last updated on 03/Mar/22

$$x^3-3.x^2.\sqrt{2}+3.x.{\left(\sqrt{2}\right)}^2-{\left(\sqrt{2}\right)}^3=8$$$$\Rightarrow {(x-\sqrt{2})}^3=2^3$$$$\Rightarrow x=2+\sqrt{2}$$$$\therefore x^5-41x^2+2022$$$$={(2+\sqrt{2})}^5-41{(2+\sqrt{2})}^2+2022$$$$=32+5\times 16\times \sqrt{2}+10\times 8\times 2+10\times 4\times 2\sqrt{2}$$$$+5\times 2\times 4+4\sqrt{2}-41(4+4\sqrt{2}+2)+2022$$$$=32+160+40+80\sqrt{2}+80\sqrt{2}+4\sqrt{2}$$$$-246-164\sqrt{2}+2022$$$$=2022-14=2008$$

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