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

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

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

$${\mathrm{x}}^3=\frac{1+3\sqrt{2004}+3\times 2004+2004\sqrt{2004}}{8}$$$${4\mathrm{x}}^3=\frac{1+3(1335+669)+2007\sqrt{2004}}{2}$$$${4\mathrm{x}}^3=\frac{4006+2007(1+\sqrt{2004})}{2}=2003+2007\mathrm{x}$$$$\therefore 4{\mathbf{x}}^3-2007\mathbf{x}-2000=3$$

Commented by jagoll last updated on 01/May/20

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

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