2022/053) If $x^2-x=1$ Symplify $\frac{x^5+8}{x+1}$

 We have $x^2 = 1-x\cdots(1)$

multiply both by x to get

$x^3 = x -x^2 = x - (1-x) = 2x -1\cdots(2)$

 $x^5 = x^3 .x x^2 = (2x-1)(1-x) = -2x^2 +3x -1 = -2(1-x) + 3x - 1 = 5x-3$ (using (1))

or $x^5 + 8 = 5x + 5$

or  $\frac{x^5+8}{x+1} = 5$