# A Not So Sinful Delight

Problem:

Show that if $x+\frac{1}{x}=1$ then $x^7+\frac{1}{x^7}=1.$

Solution-1:

From

$x+\frac{1}{x} = 1,\quad\quad\quad(0)$

we have

$x^2+1=x\quad\quad\quad(1)$

$\implies x^2=x-1\quad\quad\quad(2)$

$\implies x^3=x^2-x\overset{(2)}{=}(x-1)-x=-1\quad\quad\quad(3)$

$\implies x^6=1\quad\quad\quad(4)$

$\implies x^7=x\quad\quad\quad(5)$

$\implies x^7+\frac{1}{x^7}\overset{(5)}{=}x+\frac{1}{x}\overset{(0)}{=}1.$

Solution-2:

Exercise-1 Given $x^3+4x=8$, determine the value of $x^7+64x^2.$