[Siren-Of-Step]

ข้อ 15 ได้ว่า $4x = x^2+3x+1$
$x^2-x+1 = 0$
$x+\frac{1}{x} = 1$
$x^2+\frac{1}{x^2} = -1$
ให้ $A = \frac{4x^2}{x^4+3x^2+1} $
$\frac{1}{A} = \frac{x^4+3x^2+1}{4x^2} =\frac{x^2}{4} + \frac{1}{4x^2} + \frac{3}{4} = \frac{-1}{4}+\frac{3}{4} =\frac{2}{4} $
$A = 2$