Man This drives Me crazy!!!!!!

First, given that $x^{2} + x + 1 = 0$ , we have that

$x + \frac{1}{x}\ = \frac{x^{2} + 1}{x} = \frac{-x}{x} = -1$ .

Next, since $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$ ,

we see that $x^{2} + \frac{1}{x^{2}} = (-1)^{2} - 2 = -1$ .

Now $(x + \frac{1}{x})(x^{2} + \frac{1}{x^{2}}) = x^{3} + \frac{1}{x^{3}} + (x + \frac{1}{x})$ ,

and so $x^{3} + \frac{1}{x^{3}} = (-1)(-1) - (-1) = 2$ .

Next, $(x^2 + \frac{1}{x^{2}})^{2} = x^{4} + \frac{1}{x^{4}} + 2$ ,

and so $x^{4} + \frac{1}{x^{4}} = (-1)^{2} - 2 = -1$ .

Continuing, we have $(x^{2} + \frac{1}{x^{2}})(x^{3} + \frac{1}{x^{3}}) = x^{5} + \frac{1}{x^{5}} + (x + \frac{1}{x})$ ,

and so $x^{5} + \frac{1}{x^{5}} = (-1)(2) - (-1) = -1$ .

Now $(x^{3} + \frac{1}{x^{3}})^{2} = x^{6} + \frac{1}{x^{6}} + 2$ ,

and so $x^{6} + \frac{1}{x^{6}} = 2^{2} - 2 = 2$ .

Finally, evaluating the given expression, we have an answer of

$(-1)^{2} + (-1)^{2} + 2^2 + (-1)^{2} + (-1)^{2} + 2^{2} = \boxed{12}$ .

From the given $x^{2} +x+1=0$ we know $x=w$ OR $x=w^{2}$

( where w being cube root of unity i.e. $\frac{-1+\sqrt{3}i}{2}$ )

(and we know that $\frac{1}{w}=w^{2}$ and $\frac{1}{w^{2}}=w$ because $w^{3}=1$

and $1+w+w^{2}=0$ => $w+w^{2}=-1$

$x+\frac{1}{x}=w+\frac{1}{w}$ =-1

$x^{2}+\frac{1}{x^{2}}=w^{2}+\frac{1}{w^{2}}$ =-1

$x^{3}+\frac{1}{x^{3}}=1 + 1$ =2

[ $x^{n}=1$ whenever n is a multiple of 3 ]

So value comes = 1 + 1 +4 +1 +1 +4 = $\boxed{12}$

It is given that $x^2+x+1.$

Now ,

$x^2+x+1 =0 \Rightarrow x^2+1=-x \Rightarrow x(x+\dfrac{1}{x}) = -x \Rightarrow \boxed {x+\dfrac{1}{x}=-1}.$

Squaring on both sides,

$x^2+\dfrac{1}{x^2}+2=1 \Rightarrow \boxed{ x^2+\dfrac{1}{x^2}=-1}.$

Now cubing the first equation, we get , $x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x}) = -1 \Rightarrow \boxed{x^3+\dfrac{1}{x^3}=2}.$

Next , squaring the second equation , we get ,

$x^4+\dfrac{1}{x^4}+2 = 1 \Rightarrow \boxed{x^4+\dfrac{1}{x^4}=-1}.$

Now , by multiplying second and third equations ,

$x^5+\dfrac{1}{x^5}+x+\dfrac{1}{x}=-2 \Rightarrow \boxed{x^5+\dfrac{1}{x^5}=-1}.$

Now , by multiplying first and fifth equations ,

$x^6+\dfrac{1}{x^6}+x^4+\dfrac{1}{x^4}=1 \Rightarrow \boxed{x^6+\dfrac{1}{x^6}=2}.$

Now, the sum of their squares is

$(-1)^2+(-1)^2+(2)^2+(-1)^2+(-1)^2+(2)^2 = 1+1+4+1+1+4 = \boxed{12}.$

Solving ${x}^{2}+{x}+1$ using the quadratic formula or what not to find: $x=\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$ , this complex number can then be written in trigonometric form: $x=\cos\frac{2\pi}{3}\pm\sin\frac{2\pi}{3}$ . De Moirve's theorem then says ${(\cos\theta+i\sin\theta)}^{n}=\cos n\theta+\sin n\theta$ , hence it can be figured that ${z}^{n}+\frac{1}{{z}^{n}}=2\cos n\theta$ . Where $z$ is a complex number. Since any value of $x$ is complex, all the terms in the expression we are trying to find can be simplified to $\sum _{ k=1 }^{ 6 }{ 4\cos ^{2}{ \frac { 2k\pi }{ 3 } } }$ , which can then quite easily evaluated to 12. I found that the most interesting way to solve it :)

taking x^2 as common in the given equation we get (x+1/x)=-1, squaring this we get the first term and by opening square we get our second term squaring this again got our second term and applying binomial got our more terms the will be 1+1+4+1+1+4

x^2 + x +1=0 multiply x

x^3 + x^2 + x = 0

x^3 + x^2 + x + 1 = 1

x^3 + 0 = 1

x^3 = 1

x^3 + 1/x^3 = 2, by square

x^6 + 1/x^6 = 2

since x not equal to 0, x^2 + x +1=0 divide x

x + 1/x = -1, by square

x^2 + 1/x^2 = -1, by square

x^4 + 1/x ^4 = -1

(x^3 + 1/x^3)(x^2 +1/x^2) = -2

x^5 + 1/x^5= -1

sum of all square is 12