An algebra problem by Reynan Henry

Relevant wiki: Vieta's Formula - Higher Degrees

$\begin{aligned} x^2 - x + 5 & = 0 \\ \implies x^2 & = x - 5 \\ x^3 & = x^2 - 5x = x - 5 - 5x = -4x - 5 \end{aligned}$

Since $\alpha$ and $\beta$ are roots of $x^2 - x + 5 = 0$ , then we have:

$\begin{aligned} X & = \left(\alpha^3+\alpha^2+\alpha+1\right) \left(\beta^3+\beta^2+\beta+1\right) + 56 \\ & = \left(-4\alpha-5+\alpha-5+\alpha+1 \right) \left(-4\beta-5+\beta-5+\beta+1 \right) + 56 \\ & = \left(-2\alpha-9 \right) \left(-2\beta-9 \right) + 56 \\ & = 4{\color{#3D99F6} \alpha \beta} + 18({\color{#3D99F6}\alpha + \beta}) + 81 + 56 \quad \quad \small \color{#3D99F6} \text{By Vieta's formula }\alpha \beta = 5, \ \alpha + \beta = 1 \\ & = 4({\color{#3D99F6} 5}) + 18({\color{#3D99F6}1}) + 81 + 56 \\ & = \boxed{175} \end{aligned}$

with $x^2-x+5=0$ we can get $x^3-x^2+5x=0$ and $2x^2-2x+10=0$ . If we sum those equation we get $x^3+x^2+3x+10=0$ so $\alpha^3+\alpha^2+3\alpha+10=0$ and $\beta^3+\beta^2+3\beta+10=0$ . This means $\alpha^3+\alpha^2+\alpha+1=\alpha^3+\alpha^2+\alpha+1-(\alpha^3+\alpha^2+3\alpha+10)=-2\alpha-9$ . So the problem can be simplified to
$(2\alpha+9)(2\beta+9)=4\alpha\beta+18(\alpha+\beta)+81=4\cdot 5 + 18\cdot 1 +81 =119$ and the answer is $119+56 = 175$