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The cubic equation has one real root.Find it. $\displaystyle 8x^3-3x^2-3x-1=0$

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HINT:

We have $$9x^3=(x+1)^3\iff \left(1+\frac1x\right)^3=9$$

Observe that $x$ will be real or complex according as $\displaystyle 1+\frac1x $

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    $\begingroup$ You really enjoy (x + 1 / x) ! This leads, when feasible, to wonderful solutions. Cheers. $\endgroup$ Jan 17 '14 at 11:03
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    $\begingroup$ I think that should be $\left(1+\dfrac1x\right)^3 = 9$. $\endgroup$
    – peterwhy
    Jan 17 '14 at 12:05
  • $\begingroup$ @peterwhy, definitely, I couldn't rectify as I was on my way home. Thanks $\endgroup$ Jan 17 '14 at 14:26
  • $\begingroup$ @ClaudeLeibovici, wish you'd verified the mistake:) $\endgroup$ Jan 17 '14 at 14:39
  • $\begingroup$ @labbhattacharjee. Nooooo ! I was so impressed that I did not pay attention to that. $\endgroup$ Jan 17 '14 at 19:10

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