Cool Algebra Problem ...

$x^{2}+x+1=0$

$x^{3}=x(x^{2}+x+1)-(x^{2}+x)=0-(-1)=1$

${(x^{3}+x^{-3})}^{3}=8$

$x^2+x+1=0$

Multiply $\frac{1}{x}$ through the equation

$x+1+\frac{1}{x}=0$

$x+\frac{1}{x}=-1 \dots \bigstar$

Raise both sides of the equation to the power of $3$

$(x+\frac{1}{x})^3=(-1)^3$

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

$x^3+3(x+\frac{1}{x})+\frac{1}{x^3}=-1$

From $\bigstar$ we substitute -1

$x^3+3(-1)+\frac{1}{x^3}=-1$

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

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

$x^3+x^{-3}=2$

Raise both sides of the equation to the power of $3$ again

$x^3+x^{-3}=2^3$

$x^3+x^{-3}=\boxed{8}$

Multiplying given equation by x we get x³ = -x²-x = 1

And hence solution of required expression is 8

$x^2 + x + 1 = 0$

$(x^2 + x + 1)(x - 1) = 0$

$x^3 - 1 = 0$

$x^3 = 1$

$x^{-3} = (x^3)^{-1} = 1$

...and the rest is easy.

Using the quadratic formula yields: $x = \frac{-1}{2}\pm\frac{\sqrt{3}}{2}j$ Where ${j}$ = $\sqrt{-1}$ . Taking the positive root and using De Moivre's theorem: $x^3 = (cos120 + jsin120)^3 = cos360 +jsin360 = 1$ and taking the negative root: $x^3 = (cos240 +jsin240)^3 = cos720 + jsin720 = 1$ $\therefore\ (x^3 + x^{-3})^3 = (1+1)^3 = 2^3 = 8$

Omega of x^3 = 1

(1 + 1)^3 = 8

$\omega$ (Complex cube root of unity) is one of the root of the given equation. $y=(\omega ^{3} +\omega ^{-3})^{3}=(1+1)^{3}=\fbox{8}$

x+1/x=-1 then, x^3+1/x^3=2 and 2^3=8

given equation has roots omega and omegasqr .. so both cubes are equal to 1 and ans 8

$x^2 + x + 1 = 0$

Using the quadratic formula,

$x = \frac{-1 \pm \sqrt{3}}{2}$

$x = -\frac 1 2 \pm \frac{\sqrt{3}}{2} i$

Using Euler's equation,

$x = \exp(\pm \frac{2 \pi}{3} i$ )

Plugging this into the equation with y:

$(x^3 + x^{-3})^{3} = y$

$(\exp(\pm 2 \pi i) + \exp(\mp 2 \pi i))^{3} = y$

$(1 + 1)^{3} = y$

$(2)^{3} = y$

$y = 8$

x=(-1+rt3 i)/2; x= (-1+rt3 i)/2 so x=e^2pi/3 x^3=e^2pi = 1; x^-3=e^-2pi=1; so (x^3+x^-3)^3=(2)^3=8

(First attempt at a solution using LaTeX)

If { x }^{ 2 }+x+1=0 then (x-1)({ x }^{ 2 }+x+1)=0 but LHS = { x }^{ 3 }-1=0 (using the formula a^3-b^3, where a = x and b = 1)

so { x }^{ 3 }-1=0 { x }^{ 3 }\quad =\quad 1

Substituting into the equation with y gives us y={ ({ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ) }^{ 3 }\ y={ (1+\frac { 1 }{ 1 } ) }^{ 3 }\ y={ 2 }^{ 3 }\quad =\quad 8\

the roots of equation x2 + x+1 = 0 is w(omega) and w2(omega square).

w is also known as cube root of unity. i.e. (x^3 = 1)

w3(omega cube) = 1

w+w2 = -1

putting these values to equation we have (w^3 + w^(-3))^3

= (1 + 1)^3 = 8

and

(w^6+w^(-6))^3

=(1 + 1)^3 = 8

I did exactly what Jim PrevailLone did.

For x^2+x+1=0 => x=w, w^2 which are complex cube roots of unity hence for x=w; y=(x^3+x^3)^3(w^3+w^-3)^3
as w^3=w^-3=1 therefore
y=(w^3+w^-3)^3=(1+1)^3=8=>y=8 :)

x^2+x+1 has two solution (-1)^{1/3} and (-1)^{2/3} using the last one we get y=8 but using the first y=-8. Interesting would be rewriting and changing signs in first equation x^2 -x-1 then roots are golden ratio

The roots of

${ x }^{ 2 } + x + 1 = 0$

are $\omega$ and ${ \omega }^{ 2 }$ , which are the cube roots of unity.

Therefore, ${ x }^{ 3 }=1\quad$

$\Rightarrow \quad { ({ x }^{ 3 }+{ x }^{ -3 }) }^{ 3 } = 8$

x^2 + x +1 = 0 <=> x + 1/x = -1

y = (x^3 +1/x^3)^3 = ( x + 1/x)^3 * (x^2 + 1/x^2 - 1)^3 = (-1)^3 * ((x + 1/x)^2 - 3)^3) = - ((-1)^2 - 3)^3 = +8

(x-1)(x^2+x+1)=0 x^3 -1=0 x=1 So, (x^3 + x^-3)^3=2^3=8

at the equation : $x^2 + x + 1 = 0$ ---> divide both sides by $x$ we get : $x + 1 + \frac{1}{x} = 0$ , also $x + \frac{1}{x} = -1$ ---> [1] then multiply each side by itself , we get : $(x + \frac{1}{x})^2 = 1$ , also by expanding we get : $x^2 + 2 + (\frac{1}{x})^2 = 1$ ---> [2]

at the second equation : $(x^3 + (\frac{1}{x})^3)^3 = y$ ---> (factorize) we get : $( (x + \frac{1}{x})(x^2 - \frac{x}{x} + (\frac{1}{x})^2) )^3 = y$ then add and subtract 2 to the second brackets we get : $( (x + \frac{1}{x})(x^2 - 1 + (\frac{1}{x})^2 + 2 - 2) )^3 = y$

then by substituting from [1] & [2] we get : $( (-1)(1 - 1 -2) )^3 = y$ , then $( (-1)(-2) )^3 = y$

So : $y = 2^3$ ,

$y = \boxed{8}$