Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that 
$x^5 + \frac{1}{x^5} = 1$ as well.
 A: Notice that: $$x + \frac{1}{x} = 1 \;\implies\; x^2 - x + 1 = 0 \;\implies\; x^3 + 1 = (x+1)(x^2-x+1) = 0 \;\implies\; x^3=-1$$
Then $x^6=1$ so: $$x^5 + \frac{1}{x^5} = \frac{x^6}{x} + \frac{x}{x^6} = \frac{1}{x} + x = 1$$
A: Notice that $x+x^{-1}=1$ implies that
$$1 = (x+x^{-1})^3 = (x^3+x^{-3}) + 3(x+x^{-1}),$$
and thus
$$x^3+x^{-3} = -2.$$
Therefore,
$$1 = (x+x^{-1})^5 = (x^5+x^{-5}) + 5(x^3+x^{-3}) + 10(x+x^{-1}),$$
giving
$$x^5+x^{-5} = 1.$$
A: Since $x + \frac{1}{x} = 1$ then
\begin{align}
1 &= 1^5 = \left( x + \frac{1}{x} \right)^{5} \\
&= \left( x^5 + \frac{1}{x^5} \right) + 5 \, \left( x^3 + \frac{1}{x^3} \right) + 10 \, \left( x + \frac{1}{x} \right) \\
&= \left( x^5 + \frac{1}{x^5} \right) + 5 \, \left(x + \frac{1}{x} \right)^3 - 15 \, \left( x + \frac{1}{x} \right) + 10 \, \left( x + \frac{1}{x} \right) \\
1 &= \left( x^5 + \frac{1}{x^5} \right) + 5 \, \left(x + \frac{1}{x} \right)^3 - 5 \, \left( x + \frac{1}{x} \right) \\
1 &= x^5 + \frac{1}{x^5}
\end{align}
A: Put $u=x+\frac 1x$
and let 
$u^\boxed{\tiny n}=x^n+\frac 1{x^n}$.
By expansion it is clear that
$$u^3=u^\boxed{\tiny{3}}+3u$$
and
$$u^5=u^\boxed{\tiny5}+5u^\boxed{\tiny3}+10u$$.
Eliminating $u^\boxed{\tiny3}$ gives
$$u^{\boxed{\tiny5}}=u^5-5u^3+5u=1-5+5=1$$
A: $x+\frac 1x=1\\
(x+\frac 1x)^2=1\\
x^2 + 2 + (\frac 1x)^2 = 1\\
x^2 + (\frac 1x)^2 = -1\\
x^4 + (\frac 1x)^4 = -1\\$
$x^5+(\frac 1x)^5 = (x+\frac 1x)(x^4 - x^3(\frac 1x) + x^2(\frac 1x)^2 - x(\frac 1x)^3 + (\frac 1x)^4)\\
(x+\frac 1x)(x^4 + (\frac 1x)^4 - x^2 - (\frac 1x)^2 + 1  )\\
 ( 1 )((-1) - (-1) + 1) = 1$
