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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is

I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion of $(a+b)^2$ But how to find $\alpha^6+\beta^6$ ?

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  • $\begingroup$ Do you mean $x^2$ instead of $x^3$? $\endgroup$ – Arnaud D. Apr 28 '17 at 11:47
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    $\begingroup$ I'm sorry it was $x^2$. My printed version was a bit blurred. $\endgroup$ – H G Sur Apr 28 '17 at 11:56
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    $\begingroup$ In case you have edited your question I need to edit my answer also. $\endgroup$ – R. Singh Apr 28 '17 at 11:57
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    $\begingroup$ You are most Welcome! $\endgroup$ – R. Singh Apr 28 '17 at 12:00
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    $\begingroup$ Oddly, none of the posts below reaches the answer to your question and the one you accepted might be the most useless one to solve this. The approach suggesting to express $\alpha^6+\beta^6$ in terms of $s=\alpha+\beta$ and $p=\alpha\beta$ and then to plug the values $s=2$ and $p=4$ in the expression, works, but it is long and cumbersome. Much preferable here (and what the authors have in mind) is to note that the roots of the polynomial are $$1\pm i\sqrt3=2e^{\pm i\pi/3}$$ hence $$\alpha^6+\beta^6=(2e^{i\pi/3})^6+(2e^{-i\pi/3})^6=2^6\cdot(e^{2i\pi}+e^{-2i\pi})=2^6\cdot2=128$$ $\endgroup$ – Did May 1 '17 at 14:59
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From $\alpha^2-2\alpha+4=0$ we get that for $n\geq 0$ we have $\alpha^{n+2}-2\alpha^{n+1}+4\alpha^n=0$, and we have the same relation for $\beta$. Putting $u_n=\alpha^n+\beta^n$ and adding, we get that $u_{n+2}-2u_{n+1}+4u_n=0$ for all $n$. We have $u_0=2$, $u_1=2$, and now it is easy to compute $u_6$.

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  • $\begingroup$ Nice general answer for problems like this. +1 $\endgroup$ – AlexanderJ93 Apr 28 '17 at 12:01
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Hint: Here $\alpha^2=2\alpha -4$ and same for $\beta$, as they are the roots of the equation $x^2-2x+4=0$.

Now calculate the value of ${\alpha}^6$ and ${\beta}^6$ in terms of $\alpha^2 and \beta^2$. Then substitute their value again.

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    $\begingroup$ I think $a^2=2a-4$ $\endgroup$ – Ovi Apr 28 '17 at 12:15
  • $\begingroup$ oops... Sorry for the silly calculation mistake.. $\endgroup$ – R. Singh Apr 28 '17 at 12:48
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Use this one:

$$[(a+b)^2-2ab)][((a+b)^2-2ab)^2-3a^2b^2]$$

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Well, in general:

$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag1$$

So, for your example we can set:

  • $$\alpha=x_+=\frac{-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag2$$
  • $$\beta=x_-=\frac{-\text{b}-\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag3$$

Now, for your problem:

$$\alpha^6+\beta^6=\left(\frac{-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\right)^6+\left(\frac{-\text{b}-\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\right)^6=$$ $$\frac{\left(-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}\right)^6+\left(\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}\right)^6}{64\cdot\text{a}^6}\tag4$$

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    $\begingroup$ I find this solution the easiest. $\endgroup$ – H G Sur Apr 28 '17 at 14:56
  • $\begingroup$ @HGSur Is this some kind of macabre irony? Did you actually try to implement this suggestion? And what happened then? $\endgroup$ – Did May 1 '17 at 15:01
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    $\begingroup$ @Did, I interpreted this answer in the following way : Finding the roots using the quadratic equation and then rising it to the power and adding. It's mundane but worked for this question. $\endgroup$ – H G Sur May 3 '17 at 10:08
  • $\begingroup$ @HGSur You mean, you really performed the Herculean task of expanding the sixth powers in (4), involving complex numbers and all? Wow... And which actual value did you get, once the dust had settled? $\endgroup$ – Did May 3 '17 at 11:17
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    $\begingroup$ @Did I used this tool for finding the root and then powers : wolframalpha.com/input/… $\endgroup$ – H G Sur May 3 '17 at 11:35

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