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Question Number 9989 by konen last updated on 20/Jan/17

x−(1/x)=3 ⇒x^3 −4x^2 +2x−3=?

$$\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{3}\:\Rightarrow\mathrm{x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{3}=? \\ $$

Answered by mrW1 last updated on 20/Jan/17

x−(1/x)=3  ⇒x^2 −3x−1=0  ⇒⇒x^2 =3x+1  ⇒⇒x^3 =3x^2 +x, since x≠0    x^3 −4x^2 +2x−3  =(3x^2 +x)−4x^2 +2x−3  =−x^2 +3x−3  =−(x^2 −3x−1)−4  =−0−4  =−4

$$\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\Rightarrow{x}^{\mathrm{2}} =\mathrm{3}{x}+\mathrm{1} \\ $$$$\Rightarrow\Rightarrow{x}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} +{x},\:{since}\:{x}\neq\mathrm{0} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{3} \\ $$$$=\left(\mathrm{3}{x}^{\mathrm{2}} +{x}\right)−\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3} \\ $$$$=−{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{3} \\ $$$$=−\left({x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}\right)−\mathrm{4} \\ $$$$=−\mathrm{0}−\mathrm{4} \\ $$$$=−\mathrm{4} \\ $$

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