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Question Number 91560 by ar247 last updated on 01/May/20

x=((1+(√(2004)))/2) 4x^3 −2007x−2000=?

$${x}=\frac{\mathrm{1}+\sqrt{\mathrm{2004}}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} −\mathrm{2007}{x}−\mathrm{2000}=? \\ $$

Commented by Prithwish Sen 1 last updated on 01/May/20

x^3 =((1+3(√(2004))+3×2004+2004(√(2004)))/8) 4x^3 = ((1+3(1335+669)+2007(√(2004)))/2) 4x^3 = ((4006+2007(1+(√(2004))))/2) = 2003+2007x ∴ 4x^3 −2007x−2000 = 3

$$\mathrm{x}^{\mathrm{3}} =\frac{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{2004}}+\mathrm{3}×\mathrm{2004}+\mathrm{2004}\sqrt{\mathrm{2004}}}{\mathrm{8}} \\ $$$$\mathrm{4x}^{\mathrm{3}} \:=\:\frac{\mathrm{1}+\mathrm{3}\left(\mathrm{1335}+\mathrm{669}\right)+\mathrm{2007}\sqrt{\mathrm{2004}}}{\mathrm{2}} \\ $$$$\mathrm{4x}^{\mathrm{3}} =\:\frac{\mathrm{4006}+\mathrm{2007}\left(\mathrm{1}+\sqrt{\mathrm{2004}}\right)}{\mathrm{2}}\:=\:\mathrm{2003}+\mathrm{2007x} \\ $$$$\therefore\:\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{2007}\boldsymbol{\mathrm{x}}−\mathrm{2000}\:=\:\mathrm{3} \\ $$

Commented by jagoll last updated on 01/May/20

2x = 1+(√(2004)) 4x^2 = 2005+2(√(2004)) 8x^3 = (1+(√(2004)))(2005+2(√(2004)) ) 8x^3 =2005+2007(√(2004)) +4008 4x^3 = ((6013+2007(√(2004)))/2) 4x^3 = ((6013)/2)+2007(((2x−1)/2)) 4x^3 = ((6013−2007)/2)+2007x 4x^3 = ((4006)/2)+2007x ∴4x^3 −2007x−2000 = 2003 +2007x−2007x−2000 = 3

$$\mathrm{2}{x}\:=\:\mathrm{1}+\sqrt{\mathrm{2004}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \:=\:\mathrm{2005}+\mathrm{2}\sqrt{\mathrm{2004}} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} \:=\:\left(\mathrm{1}+\sqrt{\mathrm{2004}}\right)\left(\mathrm{2005}+\mathrm{2}\sqrt{\mathrm{2004}}\:\right) \\ $$$$\mathrm{8}{x}^{\mathrm{3}} =\mathrm{2005}+\mathrm{2007}\sqrt{\mathrm{2004}}\:+\mathrm{4008} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} \:=\:\frac{\mathrm{6013}+\mathrm{2007}\sqrt{\mathrm{2004}}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} =\:\frac{\mathrm{6013}}{\mathrm{2}}+\mathrm{2007}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{4}{x}^{\mathrm{3}} =\:\frac{\mathrm{6013}−\mathrm{2007}}{\mathrm{2}}+\mathrm{2007}{x} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} =\:\frac{\mathrm{4006}}{\mathrm{2}}+\mathrm{2007}{x} \\ $$$$\therefore\mathrm{4}{x}^{\mathrm{3}} −\mathrm{2007}{x}−\mathrm{2000}\:= \\ $$$$\mathrm{2003}\:+\mathrm{2007}{x}−\mathrm{2007}{x}−\mathrm{2000} \\ $$$$=\:\mathrm{3} \\ $$

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