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

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 $\alpha^6+\beta^6$ ?

• Do you mean $x^2$ instead of $x^3$? – Arnaud D. Apr 28 '17 at 11:47
• I'm sorry it was $x^2$. My printed version was a bit blurred. – H G Sur Apr 28 '17 at 11:56
• In case you have edited your question I need to edit my answer also. – R. Singh Apr 28 '17 at 11:57
• You are most Welcome! – R. Singh Apr 28 '17 at 12:00
• 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 $$\alpha^6+\beta^6=(2e^{i\pi/3})^6+(2e^{-i\pi/3})^6=2^6\cdot(e^{2i\pi}+e^{-2i\pi})=2^6\cdot2=128$$ – Did May 1 '17 at 14:59

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 $u_6$.

• Nice general answer for problems like this. +1 – AlexanderJ93 Apr 28 '17 at 12:01

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.

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

Use this one:

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

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:

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

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