Question

If $x=9-4\sqrt{5}$ then , find the value of $x^3-\frac{1}{x^3}$

Answer 1

We have,

$x=9-4\sqrt{5}\frac{1}{x}=9+4\sqrt{5}$

Therefore,

$x-\frac{1}{x}=-8\sqrt{5}$

$x+\frac{1}{x}=18$

Since,

$x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)=\left(-8\sqrt{5}\right)\left[{\left(x+\frac{1}{x}\right)}^2-1\right]=\left(-8\sqrt{5}\right)\left[323\right]=-2584\sqrt{5}$

Hence, this is the answer.