Totally Newton

Since $\alpha$ is a root of $x^2+x+1=0\quad \Rightarrow \alpha^2 + \alpha + 1 = 0$

$\Rightarrow \alpha^3 + \alpha^2 + \alpha = 0 \quad \Rightarrow \alpha^3 + \alpha^2 + \alpha + 1 = 1 \quad \Rightarrow \alpha^3 = 1$

$\Rightarrow \alpha^4 = \alpha \quad \Rightarrow \alpha^5 = \alpha^2 \quad \Rightarrow \alpha^6 = 1 ... \quad \Rightarrow \alpha^n = \alpha^{n \text{ mod } 3}$

Similarly, $\beta^n = \beta^{n \text{ mod } 3}$

$\Rightarrow \alpha^{2015}+\beta^{2015} = \alpha^{2}+\beta^{2} = (\alpha+\beta)^2 - 2\alpha \beta = (-1)^2- 2(1) = \boxed{-1}$

The roots of given equation are $\omega \ \& \ \omega ^{2}$

So we need to find $\omega^{2015}+\omega^{4030}$

By the properties of $\omega$

• $\omega^{2015}=\omega^{2}$

• $\omega^{4030}=\omega$

So $\omega^{2015}+\omega^{4030}=\omega^{2}+\omega$

We know $\omega$ satisfies the given equation by which $\omega^{2}+\omega=-1$

So the answer is $\large -1$

Note-: $\omega$ is the cube root of unity

$\alpha, \beta\ = \dfrac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = -\frac{1}{2} \pm i \cdot \frac{\sqrt{3}}{2} = \cos\left(\frac{2}{3}\pi \right) \pm i \cdot \sin\left(\frac{2}{3}\pi\right)$

WLOG, $\alpha = cis\left(\frac{2}{3}\pi\right), \beta = cis\left(-\frac{2}{3}\pi\right)$

$\alpha^{2015} + \beta^{2015} = cis\left(\frac{2}{3}\pi \cdot 2015 \right) + cis\left(-\frac{2}{3}\pi \cdot 2015\right) = \boxed{-1}$

Note: $cis(x) = e^{ix} = \cos(x) + i\sin(x)$

$\alpha$ and $\beta$ are the roots of $x^2+x+1$

Both the zeroes of the given equation are non real [Since, $D<0$ ]

So, $\alpha\ne 1$ and $\beta\ne 1$

Clearly, $\alpha^2+\alpha+1=0$

$\Rightarrow (\alpha-1)(\alpha^2+\alpha+1)=0$

[Since, $\alpha\ne 1$ So $(\alpha-1)\ne 0$ ]

$\Rightarrow \alpha^3=1$ Similarly, $\beta^3=1$

$\Rightarrow \alpha^{2013}=(\alpha^3)^{671}=1=(\beta^3)^{671}=\beta^{2013}$

So, $\alpha^{2015}+\beta^{2015}$ =

$(\alpha^{2013})\alpha^2+(\beta^{2013})\beta^2$

$=\alpha^2+\beta^2$

$=(\alpha+\beta)^2-2\alpha\beta$

$=(-1)^2-2(1)$ ......................[Using veita, $\alpha+\beta=-1$ and $\alpha\beta=1$ ]

So, $\alpha^{2015}+\beta^{2015}=\boxed{-1}$

$\Rightarrow$ In general, if $\alpha$ and $\beta$ are the roots of the given quadratic equation, then $\alpha^{3n}=\beta^{3n}=1$ for any integer $n$

Roots are w and w^2 Thus, w^2015=w^2 and w^4030=w So, w + w^2 = -1

In fact, Newton's Sums work quite nicely here. Newton's Sums states that, for a polynomial $a_{n}x^{n} + a_{n-1}x^{n-1} + ... + a_{0}$ : $S_{1}a_{n} + a_{n-1} = 0$ $S_{2}a_{n} + S_{1}a_{n-1} + 2a_{n-2} = 0$ $S_{3}a_{n} + S_{2}a_{n-1} + S_{1}a_{n-2} + 3a_{n-3} = 0$ and so on, where $S_{n}$ denotes the sum of the roots to the nth power. Here, we are looking for $S_{2015} = \alpha^{2015} + \beta^{2015}.$ We discover that: $1S_{1} + 1 = 0 --- S_{1} = -1$ $1S_{2} + 1S_{1} + 2 = 0 --- S_{2} = -1$ $1S_{3} + 1S_{2} + 1S_{1} + 0 = 0 --- S_{3} = 2$ $1S_{4} + 1S_{3} + 1S_{2} + 0 + 0 = 0 --- S_{4} = -1$ The cycle of -1, -1, and 2 repeats, and so our answer is $S_{2015} = -1.$

So totally Newton...LOL. Discover that the roots are cis(2π/3) and cis(4π/3) with the quadratic formula. Raise to the 2015, you get cis(2•2015(mod 6)π/3) and cis(4•2015(mod 6)π/3). Interestingly, the raising to the 2015th power maps each root onto the other.

Only negative 1 will satisfies in the given roots through the cube root of unity.....so that both equation = to zero