If x + (1/x) = 4 , what the value of ((x^6 −1)/x^3 )


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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
$x^2 +(1/x^2 ) + 2 = 16 ⇒x^2 +(1/x^2 ) = 14 ((x^6 −1)/x^3 ) = x^3 −(1/x^3 ) = (x−(1/x))(x^2 +(1/x^2 )+1) x^2 +1 = 4x ⇒x^2 −4x+1 = 0 x = ((4 ± 2(√3))/2) = 2 ± (√3) (1) x = 2 + (√( 3)) ⇒(1/x) = 2−(√3) ⇒ x−(1/x) = 2+(√3) − (2−(√3)) = 2(√3) (2) x = 2−(√3) ⇒(1/x) = 2+(√3) ⇒x−(1/x) = 2−(√3) −(2+(√3)) = −2(√3) ∴ ((x^6 −1)/x^3 ) = { ((2(√3) (15) = 30(√3))),((−2(√3) (15) = −30(√3) )) :}$
$x^{2} + \frac{1}{x^{2}} + 2 = 16 \Rightarrow x^{2} + \frac{1}{x^{2}} = 14$ $\frac{x^{6} - 1}{x^{3}} = x^{3} - \frac{1}{x^{3}} = (x - \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} + 1)$ $x^{2} + 1 = 4 x \Rightarrow x^{2} - 4 x + 1 = 0$ $x = \frac{4 \pm 2 \sqrt{3}}{2} = 2 \pm \sqrt{3}$ $(1) x = 2 + \sqrt{3} \Rightarrow \frac{1}{x} = 2 - \sqrt{3}$ $\Rightarrow x - \frac{1}{x} = 2 + \sqrt{3} - (2 - \sqrt{3}) = 2 \sqrt{3}$ $(2) x = 2 - \sqrt{3} \Rightarrow \frac{1}{x} = 2 + \sqrt{3}$ $\Rightarrow x - \frac{1}{x} = 2 - \sqrt{3} - (2 + \sqrt{3}) = - 2 \sqrt{3}$ $∴ \frac{x^{6} - 1}{x^{3}} = {\begin{cases} 2 \sqrt{3} (15) = 30 \sqrt{3} \\ - 2 \sqrt{3} (15) = - 30 \sqrt{3} \end{cases}$
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