An algebra problem by vansh gupta

Since $a$ and $b$ are roots of $x^2-6x-2=0$ , by Vieta's formula , $a+b=6$ and $ab=-2$ . And using Newton's sums method for $n\ge 3$ :

$\begin{aligned} a^n+b^n & = (a+b)\left(a^{n-1}+b^{n-1}\right) - ab\left(a^{n-2}+b^{n-2}\right) \\ \implies \alpha_n & = 6 \alpha_{n-1} + 2 \alpha_{n-2} \end{aligned}$

Therefore, we have $\begin{aligned} \frac {\alpha_{10} - 2 \alpha_8}{2\alpha_9} & = \frac {6\alpha_9 + 2 \alpha_8 - 2 \alpha_8}{2\alpha_9} = \frac {6\alpha_9}{2\alpha_9} = \boxed{3} \end{aligned}$ .

Solution:

$a$ and $b$ are the roots of the polynomial $x^{2}-6x-2$

By Vieta's formula,

$\large\color{#D61F06}{a+b=6 \hspace{1cm}\cdots(1)\\\large ab=-2\hspace{1cm}\cdots(2)}$

$\begin{aligned}\large\dfrac{\alpha_{10}-2\alpha_{8}}{2\alpha_9}&=\large\dfrac{a^{10}+b^{10}-2a^{8}-2b^{8}}{2(a^{9}+b^9)}\\ &=\large\dfrac{a^{10}+a^{9}b+b^{10}+b^9a}{2(a^9+b^9)}\hspace{1cm}\color{#3D99F6}(-2=ab)\\ &=\large\dfrac{a^9(a+b)+b^9(a+b)}{2(a^9+b^9)}\\ &=\large\dfrac{(a+b)\cancel{\color{#D61F06}(a^9+b^9)}}{2\cancel{\color{#D61F06}(a^9+b^9)}}\\ &=\large\dfrac{6}{2}\hspace{4cm}\color{#3D99F6}(a+b=6)\\ &=\large\boxed{3}\end{aligned}$

Since $a$ and $b$ are the roots of the quadratic, we have

$x^2 - 6x -2 = 0 \ \begin{cases} a^2 - 6a -2 = 0 \implies a^{10} - 6a^9 - 2a^8 = 0 & \small{\color{#3D99F6} \text{Multiplying both sides of the equation by } a^8} \\ \\ b^2 - 6b -2 = 0 \implies b^{10} - 6b^9 - 2b^8 = 0 & \small{\color{#3D99F6} \text{Multiplying both sides of the equation by } b^8} \end{cases}$

Adding both cases, we get

$\begin{aligned} & {\color{#D61F06} \left( a^{10} + b^{10} \right)} - 6 {\color{#20A900} \left( a^9 + b^9 \right)} - 2 {\color{#EC7300} \left( a^8 + b^8 \right)} = 0 \\ \implies & {\color{#D61F06} \alpha_{10}} - 6 {\color{#20A900} \alpha_9} - 2 {\color{#EC7300} \alpha_8} = 0 \\ \implies & \alpha_{10} - 2 \alpha_8 = 6 \alpha_9 \\ \implies & \dfrac{\alpha_{10} - 2 \alpha_8}{2 \alpha_9} = \boxed{3} \end{aligned}$