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If $$x+{1\over x} = r $$ then what is $$x^3+{1\over x^3}$$


$(a) 3,$

$(b) 3r,$


$(d) 0$

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Note, none of the options is correct.. – Thomas Andrews Jun 16 '13 at 15:32
@Ghost please accept an answer if you think it is satisfactory – dajoker Jul 29 '13 at 16:38

Given, $x+1/x=r$

$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\cdot x\cdot\frac1x\left(x+\frac1x\right)$

which gives us, $x^3+\frac1{x^3}$ = $ r^3-3r$

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$$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\cdot x\cdot\frac1x\left(x+\frac1x\right)$$

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I thought of that. I get $r^3 - 3r$ that way. But what am I to do next? – Ghost Jun 16 '13 at 15:27
@Ghost, probably validate the options – lab bhattacharjee Jun 16 '13 at 15:28

$$(x+{1\over x})^3 = r^3\to x^3+{1\over x^3}+3(x+{1\over x})=r^3\to x^3+{1\over x^3}=r^3-3r$$

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$\displaystyle r^3=\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3(x)\frac{1}{x}\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}+3r$

$\displaystyle \Rightarrow r^3-3r=x^3+\frac{1}{x^3}$

Your options are incorrect.For a quick counter eg. you can take $x=1/2$ to get $r=\frac{5}{2}$ and $x^3+\frac{1}{x^3}=\frac{65}{8}$ but none of the options result in $65/8$

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you can use $\\left( \\right)$ in case of multiple-term expressions – lab bhattacharjee Jun 16 '13 at 15:29

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