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Let $x\in \Bbb R$ such that $\dfrac{x}{(x^{2}+x+1)}=\dfrac{1}{4}$. What is the value of $\dfrac{x^{3}}{(x^{6}+x^{3}+1)}$?

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closed as off-topic by egreg, Grigory M, Dan Rust, M Turgeon, azimut Jan 1 '14 at 16:47

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up vote 7 down vote accepted

As $x\ne0,$ we have $$4=\frac{x^2+x+1}x=x+\frac1x+1\implies x+\frac1x=3$$

$$\text{Again }\frac{x^3}{x^6+x^3+1}=\frac1{x^3+\frac1{x^3}+1}$$

and using $\displaystyle a^3+b^3=(a+b)^3-3ab(a+b),$ $$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\cdot x\cdot\frac1x\left(x+\frac1x\right)=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)$$

Can you take it home from here?



$$\implies x^{m+n}+\frac1{x^{m+n}}=\left(x^m+\frac1{x^m}\right)\left(x^n+\frac1{x^n}\right)-\left(x^{m-n}+\frac1{x^{m-n}}\right)\ \ \ \ (1)$$

$$n=1\implies x^{m+1}+\frac1{x^{m+1}}=\left(x^m+\frac1{x^m}\right)\left(x+\frac1x\right)-\left(x^{m-1}+\frac1{x^{m-1}}\right)\ \ \ \ (2)$$

$m=1$ in $(2),$ $$\implies x^2+\frac1{x^2}=\left(x+\frac1x\right)\left(x+\frac1x\right)-\left(x^0+\frac1{x^0}\right)=\left(x+\frac1x\right)^2-2$$

$m=2$ in $(2),$ $$m=2\implies x^3+\frac1{x^3}=\left(x^2+\frac1{x^2}\right)\left(x+\frac1x\right)-\left(x+\frac1x\right)=\left(\left(x+\frac1x\right)^2-2\right)\left(x+\frac1x\right)-\left(x+\frac1x\right)=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)$$

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i think your answer is superior to mine. this is a very nice observation. i learnt something new today. – Lost1 Jan 1 '14 at 15:05
@Lost1, many a times, solution may be too complex to deal with:) For example, $$x+\frac1x=1, x^3=?$$ – lab bhattacharjee Jan 1 '14 at 15:07
yeah, if i am a teacher, then this is a manipulating surds questions for the average kids :P – Lost1 Jan 1 '14 at 15:10
@Lost1, yes $x+\frac1x$ is really fascinating term :… and… – lab bhattacharjee Jan 1 '14 at 15:26
@CarstenSchultz, not sure if I've understood your query/confusion? – lab bhattacharjee Jan 1 '14 at 15:44

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