Think Logically not mathematically [part-32]

$\begin{aligned} X & = \color{#3D99F6}{a^3}+a^2+a + \color{#3D99F6}{b^3}+b^2+b + \color{#3D99F6}{c^3}+c^2+c \quad \quad \small \color{#3D99F6}{\text{Note: 1}} \\ & = \color{#3D99F6}{1-2a}+a^2+a + \color{#3D99F6}{1-2b}+b^2+b + \color{#3D99F6}{1-2c}+c^2+c \\ & = 3-(\color{#3D99F6}{a+b+c})+\color{#D61F06}{a^2+b^2+c^2} \quad \quad \small \color{#3D99F6}{\text{Note: 2}} \\ & = 3-(\color{#3D99F6}{0})+\color{#D61F06}{(a+b+c)^2-2(ab+bc+ca)} \quad \quad \small \color{#D61F06}{\text{Note: 2}} \\ & = 3+\color{#D61F06}{(0)^2-2(2)} \\ & = \boxed{-1} \end{aligned}$

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Notes:

1. Since $a$ , $b$ and $c$ are roots of $x^3 +2x = 1$ or $x^3 = 1 - 2x$ $\implies \begin{cases} a^3 = 1-2a \\ b^3 = 1-2b \\ c^3 = 1-2c \end{cases}$

2. Vieta's formulas: $\begin{cases} a+b+c = 0 \\ ab+bc+ca = 2 \end{cases}$

By Vieta's formulas: $\begin{cases}a+b+c=0\\ab+ac+bc=2\\abc=1\end{cases}$

Therefore, $a^3+a^2+a+b^3+b^2+b+c^3+c^2+c\\=a^3+b^3+c^3+a^2+b^2+c^2+a+b+c\\=3abc+(a+b+c)((a+b+c)^2-3ab-3ac-3bc)+(a+b+c)^2-2ab-2ac-2bc+a+b+c\\=3-4\\=-1$

Given, a, b, c are the roots of the equation $x^3+2x-1=0$ $=> a+b+c=0$ $ab+bc+ca=2$ $abc=1$ and also $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ $=> a^2+b^2+c^2=0-(2\times2)=-4$ here, $a+b+c=0\space\space\space => a^3+b^3+c^3=3abc=3\times1=3\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ Therefore, $(a^3+a^2+a)+(b^3+b^2+b)+(c^3+c^2+c)\space=\space(a^3+b^3+c^3)+(a^2+b^2+c^2)+(a+b+c)$ $=3-4+0$ $=-1$