Given that $\displaystyle x^2-x-1=0$, find x^8 +$\displaystyle \frac{1}{x^8}$.
*show working and explanation PLS
Hi
$\displaystyle x^2-x-1 = 0$
$\displaystyle x^2 = x+1$
Multiplying par x
$\displaystyle x^3 = x^2+x = x+1+x = 2x+1$
Multiplying par x
$\displaystyle x^4 = 2x^2+x = 2(x+1)+x = 3x+2$
You can go on like this or use the Fibonacci sequence
If $\displaystyle x^n = a_n x+b_n$
Then $\displaystyle x^{n+1} = a_n x^2+b_n x = a_n (x+1) + b_n x = (a_n + b_n) x + a_n$
Therefore $\displaystyle a_{n+1} = a_n + b_n$ and $\displaystyle b_{n+1} = a_n$
(an) follows the Fibonacci rule
$\displaystyle x^{16} = 987x+610$
$\displaystyle x^8 = 21x+13$
$\displaystyle x^8 + \frac{1}{x^8} = \frac{987x+611}{21x+13} = 47$
Hello,
Note that $\displaystyle x \neq 0$
So you can divide $\displaystyle x^2-x-1=0$ by x on both sides, which will give :
$\displaystyle x-1-\frac 1x=0$
so $\displaystyle x-\frac 1x=1$
Now, consider $\displaystyle \left(x-\frac 1x\right)^2$
This is equal to $\displaystyle x^2+\frac{1}{x^2}-2=1$
Hence $\displaystyle x^2+\frac{1}{x^2}=3$
Square it again :
$\displaystyle 9=\left(x^2+\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^ 4}+2$
Hence $\displaystyle x^4+\frac{1}{x^4}=7$
Square it again :
$\displaystyle 49=\left(x^4+\frac{1}{x^4}\right)^2=\boxed{x^8+\fr ac{1}{x^8}}+2$
hence $\displaystyle x^8+\frac{1}{x^8}=47$