Question

If $x^2-4x-1=0$ then $x^3-\frac{1}{x^3}$=____

Answer 1

Given:$x^2-4x-1=0$

$\Rightarrow x^2-2\times 2x+4-4-1=0$

$\Rightarrow {(x-2)}^2=\sqrt{5}$

or $x=2\pm \sqrt{5}$

$\frac{1}{x}=\frac{1}{2+\sqrt{5}},\frac{1}{2-\sqrt{5}}$

$=\frac{1}{2+\sqrt{5}}\times \frac{2-\sqrt{5}}{2-\sqrt{5}}=-2+\sqrt{5},\frac{1}{2-\sqrt{5}}\times \frac{2+\sqrt{5}}{2+\sqrt{5}}=-2-\sqrt{5}$

Substituting the values of $x$ and $\frac{1}{x}$ in $x^3-\frac{1}{x^3}$ we get

$x^3-\frac{1}{x^3}$ for $x=2+\sqrt{5}$

$={(2+\sqrt{5})}^3-{(-2+\sqrt{5})}^3$

$={(2+\sqrt{5})}^3+{(2-\sqrt{5})}^3$

$=8+5\sqrt{5}+6\sqrt{5}(2+\sqrt{5})+8-5\sqrt{5}-6\sqrt{5}(2+\sqrt{5})$

$=16$

$x^3-\frac{1}{x^3}$ for $x=2-\sqrt{5}$

$={(2-\sqrt{5})}^3-{(-2-\sqrt{5})}^3$

$={(2+\sqrt{5})}^3+{(2+\sqrt{5})}^3$

$=2(8+5\sqrt{5}+6\sqrt{5}(2+\sqrt{5}))$

$=2(8+5\sqrt{5}+12\sqrt{5}+30)$

$=76+34\sqrt{5}$