Solving for $\dfrac {a^{12}+7}{a^4}$ with $a^2-a-1=0$

[Albert1]

$a\in R$ , and
$a^2-a-1=0$
find :
$\dfrac {a^{12}+7}{a^4}=?$

 

[anemone]

My solution:

Spoiler

$\begin{align*}\dfrac{a^{12}+7}{a^4}&=\dfrac{a^{12}+7(a^2-a)}{a^4}\text{since}\,\,(1=a^2-a)\\&=\dfrac{a^{12}+7a^2-7(a^4-2a^2)}{a^4}\text{since}\,\,\small a=a^2-1=(a+1)(a-1)=(a+1)(a+1-2)=a^2(a^2-2)=a^4-2a^2\\&=\dfrac{a^{12}-7a^4+21a^2}{a^4}\\&=\dfrac{a^{10}-7a^2+21}{a^2}\\&=\dfrac{a^{10}-7a^2+21(a^2-a)}{a^2}\text{again since}\,\,(1=a^2-a)\\&=\dfrac{a^{9}+14a-21}{a}\\&=\dfrac{a^{9}+14a-21(a^2-a)}{a}\\&=a^8-21a+35\\&=(21a+13)-21a+35\,\,\,\text{since $\small a^8=(a^4)^2=(3a+2)^2=9a^2+12a+4=9(a+1)+12a+4=21a+13$}\\&=48\end{align*}$

 

[Greg]

Spoiler

$a^2=a+1$, hence $a^4=3a+2$, so we have$$\frac{(3a+2)^3+7}{3a+2}=\frac{27a^3+54a^2+36a+8+7}{3a+2}$$$$=\frac{27a^2+27a+54a^2+36a+15}{3a+2}=\frac{27a+27+27a+54a+54+36a+15}{3a+2}$$$$=\frac{144a+96}{3a+2}=\frac{48(3a+2)}{3a+2}=48$$$$\text{ }$$