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If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.

Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to get $x^5 − x^3 + x − 2 = (x^2-x+1)(x^3+x^2-x-2)$ will help. We know both roots of the quadratic are complex, so we need only focus on the cubic $x^3+x^2-x-2$. How can we use this to show that the real root $a$ of it has $\lfloor a^6 \rfloor = 3$?

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  • $\begingroup$ Well the Intermediate Value Theorem says $a$ must be between $1$ and $2$. So it's certainly plausible. $\endgroup$ – Gregory Grant Feb 12 '16 at 19:47
  • $\begingroup$ Note that it is not "obvious" that it cannot be solved. Wolfram gives a closed form. $\endgroup$ – MathematicsStudent1122 Feb 12 '16 at 19:59
  • $\begingroup$ Looks like a duplicate $\endgroup$ – A.Γ. Feb 12 '16 at 19:59
  • $\begingroup$ May I ask what the source of this problem was? $\endgroup$ – David Feb 12 '16 at 20:29
  • $\begingroup$ There is a cubic formula btw... $\endgroup$ – Simply Beautiful Art Feb 12 '16 at 21:57
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We have $\sqrt[6]{3} \approx 1.2009$ and $\sqrt[6]{4} \approx 1.2599$. Let $f(x)=x^3+x^2-x-2$. Then $f(1.2) \approx -0.032$ and $f(1.25) \approx 0.2656$. So, $a$ must be between $1.2$ and $1.25$.

EDIT: As I said in the comments below, I see no way of showing that there is only one root using only precalculus. But, for completeness of my answer: If there were two roots, then the Mean Value Theorem would imply that the derivative is $0$ between the roots. But, $f'(x) = 3x^2+2x-1$. The roots of this are $x=-1$ and $x=\frac{2}{3}$. Using whichever method one likes, you can see that there is a local maximum at $x=-1$ and a local minimum at $x=\frac{2}{3}$. And, both $f(-1)$ and $f\left(\frac{2}{3}\right)$ are negative. So, there cannot be another root.

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    $\begingroup$ OP wants a synthetic proof without going into any numerical computation. $\endgroup$ – DeepSea Feb 12 '16 at 19:49
  • $\begingroup$ @Kf-Sansoo The OP doesn't mention that anywhere in their post. $\endgroup$ – Joe Johnson 126 Feb 12 '16 at 19:50
  • $\begingroup$ And there might be another root. $\endgroup$ – Hetebrij Feb 12 '16 at 19:50
  • $\begingroup$ @Hetebrij That's a good point to be mathematically complete with this question. $\endgroup$ – user19405892 Feb 12 '16 at 19:51
  • $\begingroup$ @Hetebrij I realized that as well. But, at the precalculus level, I see no rigorous way of showing it is the only one. $\endgroup$ – Joe Johnson 126 Feb 12 '16 at 19:52
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Let $f(x)=x^5-x^3+x-2$.

  • First step. Differenciate the polynomial to obtain: $$f'(x)=5x^4-3x^2+1$$ which has no zeros. Hence, $f$ has exactly one root.

  • Second step. We have that $f(1)=-3<0$ and $f(2)=24>0$. Then the root lies in the interval $(1,2)$.

  • Third step. Let $a$ be the root of $f$. We have $$a^6=a\cdot a^5=a(a^3-a+2)$$ Now define $g(x)=x(x^3-x+2)$. Also differentating, we can see easily that $g$ is increasing in $[1,2]$, so now you have to try more values on $f$, to get finer bounds for the root. If we have that $l<a<u$ and $\lfloor g(l)\rfloor=\lfloor g(u)\rfloor$, then $\lfloor a^6 \rfloor$ is also the same.

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  • $\begingroup$ A better third step: $a^6=2a-1+\frac{2}{a}$ (see this answer). The derivative of the RHS is positive on $[1,2]$, so it is growing there from 3 to 4. $\endgroup$ – A.Γ. Feb 12 '16 at 21:16
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Opposed to all of these answers, I will simply go ahead and solve the cubic:

$$0=x^3+x^2-x-2$$

Applying the cubic formula, I get:

$$x=\frac16\left(-2+\sqrt[3]{4}(\sqrt[3]{42+3\sqrt{177}})+\sqrt[3]{42-3\sqrt{177}})\right)$$

According to Wolfram|Alpha.

Then I guess you would either do the floor function after getting it in decimal form or you could somehow take the floor function now.

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