Easy
Repeatedly "add 2 and take a square root" to climb down the ladder x 8 + 1 x 8 → x 4 + 1 x 4 → x 2 + 1 x 2 → x + 1 x x^8+\tfrac1{x^8}\to x^4+\tfrac1{x^4}\to x^2+\tfrac1{x^2}\to x+\tfrac1x x 8 + x 8 1 → x 4 + x 4 1 → x 2 + x 2 1 → x + x 1 x 9 + 1 x 9 x^9+\tfrac1{x^9} x 9 + x 9 1 ( t + 1 t ) 3 = t 3 + 1 t 3 + 3 ( t + 1 t ) \left(t+\tfrac1t\right)^3=t^3+\tfrac1{t^3}+3\left(t+\tfrac1t\right) ( t + t 1 ) 3 = t 3 + t 3 1 + 3 ( t + t 1 )
1
Climb down to x 4 + 1 x 4 x^4+\tfrac1{x^4} x 4 + x 4 1 Add 2 2 2
x 8 + 1 x 8 + 2 = 47 + 2 = 49 ⇒ ( x 4 + 1 x 4 ) 2 = 49 ⇒ x 4 + 1 x 4 = 7 ( positive, as x > 0 ) \begin{aligned}
&x^8+\frac1{x^8}+2=47+2=49\\
&\Rightarrow\ \left(x^4+\frac1{x^4}\right)^2=49\\
&\Rightarrow\ x^4+\frac1{x^4}=7 \quad(\text{positive, as } x\gt 0)
\end{aligned} x 8 + x 8 1 + 2 = 47 + 2 = 49 ⇒ ( x 4 + x 4 1 ) 2 = 49 ⇒ x 4 + x 4 1 = 7 ( positive, as x > 0 ) 2
Down to x 2 + 1 x 2 x^2+\tfrac1{x^2} x 2 + x 2 1 x + 1 x x+\tfrac1x x + x 1
( x 2 + 1 x 2 ) 2 = x 4 + 1 x 4 + 2 = 7 + 2 = 9 ⇒ x 2 + 1 x 2 = 3 ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 = 3 + 2 = 5 ⇒ x + 1 x = 5 \begin{aligned}
&\left(x^2+\frac1{x^2}\right)^2=x^4+\frac1{x^4}+2=7+2=9\ \Rightarrow\ x^2+\frac1{x^2}=3\\
&\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2=3+2=5\ \Rightarrow\ x+\frac1x=\sqrt5
\end{aligned} ( x 2 + x 2 1 ) 2 = x 4 + x 4 1 + 2 = 7 + 2 = 9 ⇒ x 2 + x 2 1 = 3 ( x + x 1 ) 2 = x 2 + x 2 1 + 2 = 3 + 2 = 5 ⇒ x + x 1 = 5 3
Cube up to x 3 + 1 x 3 x^3+\tfrac1{x^3} x 3 + x 3 1 using t 3 + 1 t 3 = ( t + 1 t ) 3 − 3 ( t + 1 t ) t^3+\tfrac1{t^3}=\left(t+\tfrac1t\right)^3-3\left(t+\tfrac1t\right) t 3 + t 3 1 = ( t + t 1 ) 3 − 3 ( t + t 1 )
x 3 + 1 x 3 = ( 5 ) 3 − 3 5 = 5 5 − 3 5 = 2 5 x^3+\frac1{x^3}=(\sqrt5)^3-3\sqrt5=5\sqrt5-3\sqrt5=2\sqrt5 x 3 + x 3 1 = ( 5 ) 3 − 3 5 = 5 5 − 3 5 = 2 5 4
Cube once more with t = x 3 t=x^3 t = x 3 x 9 + 1 x 9 x^9+\tfrac1{x^9} x 9 + x 9 1
x 9 + 1 x 9 = ( x 3 + 1 x 3 ) 3 − 3 ( x 3 + 1 x 3 ) = ( 2 5 ) 3 − 3 ( 2 5 ) = 40 5 − 6 5 = 34 5 \begin{aligned}
x^9+\frac1{x^9}&=\left(x^3+\frac1{x^3}\right)^3-3\left(x^3+\frac1{x^3}\right)\\
&=(2\sqrt5)^3-3(2\sqrt5)\\
&=40\sqrt5-6\sqrt5=34\sqrt5
\end{aligned} x 9 + x 9 1 = ( x 3 + x 3 1 ) 3 − 3 ( x 3 + x 3 1 ) = ( 2 5 ) 3 − 3 ( 2 5 ) = 40 5 − 6 5 = 34 5
x 9 + 1 x 9 = 34 5 x^9+\dfrac1{x^9}=\mathbf{34\sqrt5} x 9 + x 9 1 = 34 5
The handy identity throughout: ( t + 1 t ) 2 = t 2 + 1 t 2 + 2 \left(t+\tfrac1t\right)^2=t^2+\tfrac1{t^2}+2 ( t + t 1 ) 2 = t 2 + t 2 1 + 2 ( t + 1 t ) 3 = t 3 + 1 t 3 + 3 ( t + 1 t ) \left(t+\tfrac1t\right)^3=t^3+\tfrac1{t^3}+3\left(t+\tfrac1t\right) ( t + t 1 ) 3 = t 3 + t 3 1 + 3 ( t + t 1 )