The story of 3 cubes

If $\alpha,\beta,\gamma$ are the roots of the equation $x^3+2x^2-3x-2=0$ , then

$\alpha^3+2\alpha^2-3\alpha-2=0$ , $\rightarrow$ $(i)$

$\beta^3+2\beta^2-3\beta-2=0$ , $\rightarrow$ $(ii)$

$\gamma^3+2\gamma^2-3\gamma-2=0$ , $\rightarrow$ $(iii)$

Multiplying $(i)$ by $\alpha^2$ , $(ii)$ by $\beta^2$ and $(iii)$ by $\gamma^2$ , we get,

$\alpha^5+2\alpha^4-3\color{#D61F06}\alpha^3-\color{#20A900}2\alpha^2$ $=0$

$\beta^5+2\beta^4-3\color{#D61F06}\beta^3-\color{#20A900}2\beta^2$ $=0$

$\gamma^5+2\gamma^4-3\color{#D61F06}\gamma^3-\color{#20A900}2\gamma^2$ $=0$

Adding all the 3 equations above, we get,

$(\alpha^5+\beta^5+\gamma^5)+2(\alpha^4+\beta^4+\gamma^4)-\color{#D61F06}3(\alpha^3+\beta^3+\gamma^3)-\color{#20A900}2(\alpha^2+\beta^2+\gamma^2)$ $=0$

$\implies$ $(\alpha^5+\beta^5+\gamma^5)+2(\alpha^4+\beta^4+\gamma^4)=\color{#D61F06}3(\alpha^3+\beta^3+\gamma^3)+\color{#20A900}2(\alpha^2+\beta^2+\gamma^2)$

We require $(\alpha^5+\beta^5+\gamma^5)+2(\alpha^4+\beta^4+\gamma^4)+\color{#D61F06}(\alpha^3+\beta^3+\gamma^3)$ ,

So we add $\color{#D61F06}(\alpha^3+\beta^3+\gamma^3)$ on both sides of the above equation, resulting,

$(\alpha^5+\beta^5+\gamma^5)+2(\alpha^4+\beta^4+\gamma^4)+\color{#D61F06}(\alpha^3+\beta^3+\gamma^3)=4(\alpha^3+\beta^3+\gamma^3)+\color{#20A900}2(\alpha^2+\beta^2+\gamma^2)$ $\rightarrow$ $(iv)$

$\rightarrow$ $\color{#EC7300}\alpha+\beta+\gamma$ $=-2$

$\rightarrow$ $\color{#E81990}\alpha\beta+\beta\gamma+\gamma\alpha=-3$

$\rightarrow$ $\color{#3D99F6}\alpha\beta\gamma=2$

$\implies$ $\color{#20A900}(\alpha^2+\beta^2+\gamma^2)$ $=\color{#EC7300}(\alpha+\beta+\gamma)^2 -\color{#E81990}2(\alpha\beta+\beta\gamma+\gamma\alpha)$ $=10$ $\rightarrow$ $(v)$

$\implies$ $\color{#D61F06}(\alpha^3+\beta^3+\gamma^3)$ $=\color{#EC7300}(\alpha+\beta+\gamma)[\color{#20A900}(\alpha^2+\beta^2+\gamma^2)-$ $\color{#E81990}(\alpha\beta+\beta\gamma+\gamma\alpha)]+\color{#3D99F6}3\alpha\beta\gamma$ $=-20$ $\rightarrow$ $(vi)$

Substituting $(v)$ and $(vi)$ in $(iv)$ ,

We obtain the final answer

$\implies$ $(\alpha^5+\beta^5+\gamma^5)+2(\alpha^4+\beta^4+\gamma^4)+\color{#D61F06}(\alpha^3+\beta^3+\gamma^3)$ $=\color{#D61F06}4(-20)+\color{#20A900}2(10)=\color{#69047E}\boxed{-60}$

$\alpha,\beta,\gamma$ are the roots of the equation $x^3+2x^2-3x-2=0$

We have,

$\begin{aligned} x^3&=2+3x-2x^2\hspace{5mm}\color{#3D99F6}(1)\end{aligned}$

Also,

$\begin{aligned}x(x^2+2x-3)&=2\\ (x+1)^2-4&=\dfrac{2}{x}\hspace{3mm}\color{#3D99F6} x\neq0\\ (x+1)^2&=4+\dfrac{2}{x}\hspace{5mm}\color{#3D99F6}(2)\end{aligned}$

Since $\alpha,\beta,\gamma$ are the roots of the equation they must satisfy $\color{#3D99F6}(1)$ & $\color{#3D99F6}(2)$ as well.

$\color{#3D99F6}(1)\times\color{#3D99F6}(2)$ gives,

$\begin{aligned} x^3\times (x+1)^2&=(2+3x-2x^2)\times(4+\dfrac{2}{x})\\ &=(\dfrac{4}{x}+14+8x-8x^2)\\ \implies \sum_{cyc \hspace{2mm} \alpha,\beta,\gamma} x^3\times (x+1)^2&=\sum_{cyc \hspace{2mm} \alpha,\beta,\gamma}(\dfrac{4}{x}+14+8x-8x^2)\hspace{5mm}\color{#3D99F6}(3)\end{aligned}$

We have,

$\begin{aligned}\alpha+\beta+\gamma&=-2\\ \alpha\beta+\beta\gamma+\gamma\alpha&=-3\\ \alpha^2+\beta^2+\gamma^2&=(\alpha+\beta+\gamma)^2-2\times(\alpha\beta+\beta\gamma+\gamma\alpha)=10\\ \alpha\beta\gamma&=2\end{aligned}$

we can note that

$\begin{aligned}\sum_{cyc \hspace{2mm} \alpha,\beta,\gamma}\dfrac{2}{\alpha}= \sum_{cyc \hspace{2mm} \alpha,\beta,\gamma}\beta\gamma\end{aligned}$

Substituting these values in $\color{#3D99F6}(3)$ we get,

$\begin{aligned}\sum_{cyc \hspace{2mm} \alpha,\beta,\gamma} x^3\times (x+1)^2&=-6+42+-16+-80=\color{#EC7300}\boxed{\color{#333333}-60}\end{aligned}$