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If $x$ is a real number that satisfies $x^{3}-\frac{1}{x^{3}}=76$, determine the value of $x-\frac{1}{x}$.

How can I solve this problem?

Josh B.
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user569622
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  • See also https://math.stackexchange.com/questions/678650/if-x-frac1x-5-find-x5-frac1x5 – lhf Jun 14 '18 at 01:40

3 Answers3

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Hint: Expand $\left(x-\dfrac{1}{x}\right)^3$.

lhf
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  • I did, and the result was $x^{3}-3\left(x-x^{-1}\right)-x^{3}$. But then I got stuck, what am I supposed to do next? – user569622 Jun 14 '18 at 01:33
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    @user569622 There is a mistake in what you wrote above. You have $$\left(x-\frac{1}{x}\right)^3 = x^3-\frac{1}{x^3} - 3\left(x-\frac{1}{x}\right)$$ Writing $X = x-\frac{1}{x}$, this means $$X^3 = 76 - 3X$$ – Clement C. Jun 14 '18 at 01:38
  • @ClementC. Thank you, I didn't realize I can write the equation like that. :) – user569622 Jun 14 '18 at 10:12
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Let $u=x-\frac {1}{x} $

$$ u^3 = 76-3u$$

$$(u-4)(u^2+4u+19)=0$$

$$u=4$$

$$x- \frac {1}{x} =4$$

Mohammad Riazi-Kermani
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$a^3-b^3=(a-b)(a^2+ab+b^2)$ and note that $(x-\frac{1}{x})^2 =x^2+\frac{1}{x^2}-2$ then if we set $(x-\frac{1}{x})=t$ we have $t(t^2+3)=76$ now find $t$.

amir bahadory
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