If-x-1-x-4-what-the-value-of-x-6-1-x-3-

Question Number 90716 by jagoll last updated on 25/Apr/20

I f x + \frac{1}{x} = 4, w h a t t h e

v a l u e o f \frac{x^{6} - 1}{x^{3}}

Commented by john santu last updated on 25/Apr/20

x^{2} + \frac{1}{x^{2}} + 2 = 16 \Rightarrow x^{2} + \frac{1}{x^{2}} = 14

\frac{x^{6} - 1}{x^{3}} = x^{3} - \frac{1}{x^{3}} = (x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} + 1)

x^{2} + 1 = 4 x \Rightarrow x^{2} - 4 x + 1 = 0

x = \frac{4 \pm 2 \sqrt{3}}{2} = 2 \pm \sqrt{3}

(1) x = 2 + \sqrt{3} \Rightarrow \frac{1}{x} = 2 - \sqrt{3}

\Rightarrow x - \frac{1}{x} = 2 + \sqrt{3} - (2 - \sqrt{3}) = 2 \sqrt{3}

(2) x = 2 - \sqrt{3} \Rightarrow \frac{1}{x} = 2 + \sqrt{3}

\Rightarrow x - \frac{1}{x} = 2 - \sqrt{3} - (2 + \sqrt{3}) = - 2 \sqrt{3}

∴ \frac{x^{6} - 1}{x^{3}} = {\begin{cases} 2 \sqrt{3} (15) = 30 \sqrt{3} \\ - 2 \sqrt{3} (15) = - 30 \sqrt{3} \end{cases}