Polynomial Scramble # 1

** It does not look like we are going to divide that mess so we use Geometric series : \color[rgb]{0.35,0.35,0.35} \dfrac{(a^8 + a^4 + 1)(a^3 + a^2 + a + 1)}{a^9 + a^6 + a^3 + 1} = 21. = $\frac{a^{12} - 1}{a^4-1} \cdot \frac{a^4-1}{a-1}/(a^{12}-1)(a^3-1)$ $= \frac{a^{12}-1}{a-1} \cdot \frac{a^3-1}{a^{12}-1}$ $= \frac{a^3-1}{a-1} = a^2 + a + 1$ . So $a^2 + a + 1 = 21$ $a^2 + a - 20$ . $a = -5 , a = 4$

Relevant wiki: Rational Root Theorem - Basic

$\begin{aligned} \frac {(a^8+a^4+1)(a^3+a^2+a+1)}{a^9+a^6+a^3+1} & = 21 \\ \frac {(a^8+a^4+1){\color{#D61F06}(a-1)}(a^3+a^2+a+1)}{{\color{#D61F06}(a-1)}(a^9+a^6+a^3+1)} & = 21 \\ \frac {(a^8+a^4+1)(a^4-1)\color{#D61F06}(a^3-1)}{(a-1){\color{#D61F06}(a^3-1)}(a^9+a^6+a^3+1)} & = 21 \\ \frac {(a^{12}-1)(a^3-1)}{(a-1)(a^{12}-1)} & = 21 \\ \frac {a^3-1}{a-1} & = 21 \\ \implies a^3 - 21a + 20 & = 0 & \small \color{#3D99F6} \text{By rational root theorem} \\ (a-1)(a^2+a-20) & = 0 \\ (a-1)(a-4)(a+5) & = 0 \\ \implies a & = \boxed{-5,4} \end{aligned}$

Note that when $a=1$ , $\dfrac {(a^8+a^4+1)(a^3+a^2+a+1)}{a^9+a^6+a^3+1} \ne 21$