Third Powers Based On Second Degree Equation?

Relevant wiki: Algebraic Manipulation Problem Solving - Basic

$x^2+1=x\implies\color{#D61F06}{\boxed{x^2=x-1}}\\ x^2=x-1\implies x^3=x^2-x\implies x^3=(x-1)-x=-1\\ \color{#3D99F6}{\boxed{x^3=-1}}\\ \implies a^3=-1,b^3=-1\implies \boxed{a^3+b^3=-2}$

$a^3 + b^3 = (a+b)(a^2-ab+b^2) \\ \text{Given equation can be written as:} \\ x^2-x+1 = 0 \\ \implies \text{Sum of roots a + b = } \dfrac{-b}{a} = \dfrac{-(-1)}{1} = 1 ... (i)\\ \text{and, product of roots ab = } \dfrac{c}{a} = \dfrac{1}{1} = 1 ... (ii) \\ \implies a^3 + b^3 = (a+b)(a^2 -ab + b^2 + 2ab - 2ab) \text{ (Add ans subtract 2ab)} \\ \implies a^3 + b^3 = (a+b)((a+b)^2 - 3ab) \\ \implies a^3 + b^3 = (1)((1)^2 - 3(1)) \text{ (From (i) and (ii))} \\ \implies a^3 + b^3 = 1(1-3) \\ \implies a^3 + b^3 = -2$

Relevant wiki: Complex Numbers

$x^{ 2 }-x+1=0\quad \rightarrow \quad \begin{cases} { \alpha }_{ 1 }=\frac { 1+\sqrt { 3 } i }{ 2 } \\ { \alpha }_{ 2 }=\frac { 1-\sqrt { 3 } i }{ 2 } \end{cases}\\ { \alpha }_{ 1 }^{ \quad 3 }=\frac { -8 }{ 8 } \quad \quad \quad \quad { \alpha }_{ 2 }^{ \quad 3 }=\frac { -8 }{ 8 } \\ { \quad \quad \quad \quad \alpha }_{ 1 }^{ \quad 3 }+{ \alpha }_{ 2 }^{ \quad 3 }=-2$

From Vieta's formula, we have $a + b = 1$ and $ab = 1$ Thus $(a + b)^3 = 1^3 = 1$ . Expanding the left hand side gives $a^3 + b^3 + 3ab (a + b) = a^3 + b^3 +3 = 1$ Hence $a^3 + b^3 = 1 - 3 = -2$

X^2 - X + 1 = f (X) f (X) = (X-a)(X-b) where a and b are complex roots of this equation a = -w, b=-w^2 where w and w^2 are cube roots of unity w = cis(2pi/3) By demorvis theorum Cis(2pi)+Cis(4pi) = 2