Cube Root Unity

${ x }^{ 3 }\quad -\quad 1\quad =\quad \left( x\quad -\quad 1 \right) \left( { x }^{ 2 }\quad +\quad x\quad +\quad 1 \right) \quad =\quad 0\\ x\quad -\quad 1\quad =\quad 0\quad \therefore \quad x\quad =\quad 1\\ { x }^{ 2 }\quad +\quad x\quad +\quad 1\quad =\quad 0\\ x\quad =\quad \frac { -1\quad \pm \quad \sqrt { { 1 }^{ 2 }\quad -\quad 4\left( 1 \right) \left( 1 \right) } }{ 2\left( 1 \right) } \quad =\quad -\frac { 1 }{ 2 } \pm \frac { \sqrt { 3 } }{ 2 } i$

Sum of roots: $1\quad +\quad \left( -\frac { 1 }{ 2 } \quad +\quad \frac { \sqrt { 3 } }{ 2 } i \right) \quad +\quad \left( -\frac { 1 }{ 2 } \quad -\quad \frac { \sqrt { 3 } }{ 2 } i \right) \\ \Longrightarrow \quad 1\quad -\quad \frac { 1 }{ 2 } \quad -\quad \frac { 1 }{ 2 } \quad +\quad \frac { \sqrt { 3 } }{ 2 } i\quad -\quad \frac { \sqrt { 3 } }{ 2 } i\quad =\quad 0\\ \Longrightarrow \quad \therefore \quad r\quad =\quad 0,\quad 1\quad +\quad { 0 }^{ 2 }\quad +\quad { 0 }^{ 3 }\quad +\quad { 0 }^{ 4 }\quad +\quad ...\quad =\quad \boxed { 1 }$

By Vieta`s formula ,we have $\color{#D61F06}{a_0+a_1+a_2=0}$ where $\color{#3D99F6}{a_0,a_1,a_2}$ are the roots of the equation. $\color{#69047E}{1+r^2+r^3+r^4\dotsm}$ can be written as: $\color{#20A900}{(1+r+r^2+r^3+\dotsm)-r}$ The first infinite series can be written as $\dfrac{1}{1-r}$ .But we know that $r=0$ so the value is: $\color{#3D99F6}{\frac{1}{1-0}-0=\boxed{1}}$

Using Vieta's formula, we know that the sum of the roots is the coefficient of x^2 times -1. Since the coefficient of x^2 is 0, r=0, which means 1 + r^2 + r^3 + r^4 + ... = 1 + 0 + 0 + 0 + ... = 1.

$1+r^2+r^3+r^4+...=\frac1{1-r}-r$ ; [ using reversed Binomial Series Expansion ]

Recalling general knowledge, $1+\omega +\omega ^2=0$ , or applying Vieta's formula , we get, $r=\sum x=0$

As such, the answer is, $\frac1{1-0}-0=1$ . QED

(This may seem an unnecessary elongation. But, this amends a rather generalized solution for more problems. For instance, I may offer this .)

x^{3} - 1=0 -----> (x-1)(x^2 + x +1) ------> r = 0. Therefore, 1 + r^{2} + r^{3} + r^{4}............ = 1answer

By Vieta's formulas, the linear term has coefficient zero, hence the desired sum is equal to one.

FOR ANY EXPRESSION OF THE FORM A(x)^n + B(x)^(n-1)+.................+N(x)+M=0 SUM OF THE ROOTS = -B/A So, Here r=0 Therefore answer is 1

since the sum of the roots of x^3-1 is 0,hence ,r is 0. Therefore,ans is 1

You can see that the roots of the problem is either 1 or -1, so that makes r=0. Therefore, if you insert that into the question, you will get 1.

One way to go about solving this problem is to use complex exponentials. If we extend this problem to the complex plane:

$|z|^{3} = 1 = e^{i 2 \pi n}$

$\Rightarrow z = e^{\frac{i 2 \pi n}{3} }$ where n is any integer. In the complex plane, the three roots are separated by $2\pi/3$ radians. We can then consider the n = 0, 1, 2 cases since the rest are just the same three roots cycled over. Doing so we obtain that the three roots of this equation are $z_0 = 1, z_1 = e^{\frac{i2\pi}{3}}, z_2 = e^{\frac{i4\pi}{3}}$ Adding them together: $r = 1 + \left[cos\left(\frac{2\pi}{3}\right) + isin\left(\frac{2\pi}{3}\right)\right] + \left[cos\left(\frac{4\pi}{3}\right) + isin\left(\frac{4\pi}{3}\right) \right]$

$= 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} - \frac{1}{2} - i\frac{\sqrt{3}}{2} = 0$

And so the value of:

$1 + r^2 + r^3 + r^4 + ... = 1 + 0 + 0 + ... = \boxed{1}$

The solution of x^3-1=0 are nothing but the cube root of unity. The sum of the cube root of unity is zero So r=0. So 1+r+r2+r^3+ etc=1

r is equal to x^3 - 1 which we are told equals 0. 0 to the power of any number is 0, therefore we are left with 1.

According to the sum of the roots will be 'r', which is: x^3-1=0. Therefore, r=0. After that we'll solve to the following sum: 1+r^2+r^3+r^4+....=?; 1+0^2+0^3+0^4+.... =1+0+0+0+0..…=1. Therefore, 1is the answer.

Xcube=1 from here we can use the method of sum of cube root of unity(nth roots of unity) which say's 1+omega+omega square=0 Where omega conjucate=omega square Omega =-1+√3i/2 Omega square= -1-√3i/2

since, 1+w+w^2=0, therefore r=0, and therefore the ans is 1

Just apply complex number theory. x^3=1.Thus there are three roots of 'x' which are the cube roots of unity:1,w,w^2. And we know,1+w+w^2=0. So,r=0. Ans:1