If x2−3x+1=0,x^2 - 3x +1 = 0, x2−3x+1=0, then one of the values of x2−8x−1x3+1/x\frac{x^2-8x-1}{x^3+1/x}x3+1/xx2−8x−1is: A5+87\frac{\sqrt5+8}{7}75+8 B−7/3-7/3−7/3 C-3 D5−87\frac{\sqrt5-8}{7}75−8 Show answer & explanationCorrect answer: DExplanationThe correct answer is Option 4 i.e. 5−87\frac{\sqrt5-8}{7}75−8. x2−3x+1=0\quad x^2 - 3x + 1 = 0 x2−3x+1=0 ⇒x+1x=3⇒x2+1x2=(x+1x)2−2=9−2=7⇒x−1x=(x+1x)2−4x×1x=9−4=5\Rightarrow x + \frac{1}{x} = 3 \\ \Rightarrow x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = 9 - 2 = 7 \\ \Rightarrow x - \frac{1}{x} = \sqrt{(x + \frac{1}{x})^2 - 4x \times \frac{1}{x}} = \sqrt{9 - 4} = \sqrt{5} \\ ⇒x+x1=3⇒x2+x21=(x+x1)2−2=9−2=7⇒x−x1=(x+x1)2−4x×x1=9−4=5 x2−8x−1x3+1x=x−8−1xx2+1x2=5−87\quad \frac{x^2 - 8x - 1}{x^3 + \frac{1}{x}} \\ = \frac{x - 8 - \frac{1}{x}}{x^2 + \frac{1}{x^2}} = \frac{\sqrt{5} - 8}{7} \\x3+x1x2−8x−1=x2+x21x−8−x1=75−8