If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is

Question

I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$

I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?

Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :)

Answer 1

$$(-1)^3=(x^4+x^2)^3=(x^3)^4+(x^3)^2+3(x^3)^2(-1)$$

Set $x^3=y$

$$y^4-2y^2+1=0$$ whose roots are $\alpha^3$ etc.

$$(y^4-2y^2+1)^2=0$$

Set $y^2=z$

$$z^4-4z^3+\cdots+1=0$$ whose roots are $(\alpha^3)^2$ etc.

Now apply Vieta's formula

Answer 2

$$x^4=-(x^2+1)\Rightarrow x^6=-x^4-x^2=x^2+1-x^2=1$$

So $$\sum x^6=\sum 1=1+1+1+1=4$$

Where $\displaystyle \sum x^6=\alpha^6+\beta^6+\gamma^6+\delta^6.$

Answer 3

Hint: Recall that $(a-b)(a^2+ab+b^2) = a^3-b^3$

Multiplying $$x^4 + x^2 + 1 = 0$$ with $x^2-1$ we get $$x^6-1=0$$ so $x^6=1$ and now is easy to finish...

Answer 4

Setting $t=x^2$, $t$ satisfies the quadratic equation $\;t^2+t+1=0$, and it is well known that the roots of this equation are the non-real cubic roots ot unity $$j=\mathrm e^{\tfrac{2i\pi}3}, \qquad \bar j=\mathrm e^{-\tfrac{2i\pi}3}.$$ Therefore the roots of the initial equation are the square roots of the previous roots, i.e. $$x\in\biggl\{\pm\mathrm e^{\tfrac{i\pi}3},\:\pm\mathrm e^{\tfrac{-i\pi}3}\biggr\},$$

and in each case, you can easily see that $x^6=1$.

Answer 5

If you are going the hard way of finding explicitly the roots, then the roots and their exponential forms are $$\alpha=-\frac{1}{2}-\frac{i\sqrt{3}}{2}=-e^{\frac{i\pi}{3}} \Rightarrow\alpha^6=e^{i2\pi}=1$$ $$\beta=\frac{1}{2}+\frac{i\sqrt{3}}{2}=e^{\frac{i\pi}{3}}\Rightarrow\beta^6=e^{i2\pi}=1$$ $$\gamma=\frac{1}{2}-\frac{i\sqrt{3}}{2}=-e^{\frac{i2\pi}{3}}\Rightarrow\gamma^6=e^{i4\pi}=1$$ $$\delta=-\frac{1}{2}+\frac{i\sqrt{3}}{2}=e^{\frac{i2\pi}{3}}\Rightarrow\delta^6=e^{i4\pi}=1$$

Answer 6

$$ (x^2-1)(x^4+x^2+1) = x^6 - 1 $$ $x^6-1=0$ if and only if $x$ is one of the $6$th roots of $1.$

$(x^2-1)(x^4+x^2+1)=0$ if and only if either $x^2-1=0,$ in which case $x=\pm 1,$ or $x^4+x^2+1=0,$ which fails to hold if $x=\pm 1,$ since $(\pm 1)^4+(\pm 1)^2 + 1 = 3\ne0.$ Thus the roots of $x^4+x^2+1=0$ are the $6$th roots of $1$ other than $\pm 1.$ Their $6$th powers are equal to $1,$ so the sum of their $6$th powers is $4.$

Answer 7

An alternative approach. Your polynomial is $$ x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1)(x^2+x+1) = \Phi_3(x)\Phi_6(x) $$ so for any root $\xi$ of the LHS, $\xi^6=1$. This implies $p_6=4$.