SSC CGL 16 Aug 2015 Shift 1 Paper

${\mathit{x}}^8$ – 1 = ${\left({\mathit{x}}^4\right)}^2$ – 12
= (${\mathit{x}}^4$ + 1) (${\mathit{x}}^4$ – 1)
= (${\mathit{x}}^4$ + 1) (${\mathit{x}}^2$ + 1) (${\mathit{x}}^2$ – 1)
= (${\mathit{x}}^4$ + 1) (${\mathit{x}}^2$ + 1) (x + 1) (x – 1)
[Since ${\mathit{a}}^2\mathit{–}{\mathit{b}}^2$ = (a + b) (a – b)]
${\mathit{x}}^4+2{\mathit{x}}^3$ – 2x – 1
= (${\mathit{x}}^4$ – 1) + $2{\mathit{x}}^3$ – 2x
= (${\mathit{x}}^2$ + 1) (${\mathit{x}}^2$ – 1) + 2x (${\mathit{x}}^2$ – 1)
= (${\mathit{x}}^2$ + 1 + 2x) (${\mathit{x}}^2$ – 1)
= ${\left(\mathit{x}+1\right)}^2$ (x + 1) (x – 1)
∴ H.C.F = (x + 1) (x – 1)
= ${\mathit{x}}^2$ – 1