Question

If $x^2-4x-5=0$ then $x^3-\frac{1}{x^3}=$

Answer 1

given, $x^2-4x-5=0$

$x^2-5x+x-5=0$

$x(x-5)-1(x-5)=0$

$(x-5)(x-1)=0$

$x=1,5$

Hence, for $x=1$,

$x^3-\frac{1}{x^3}=1-1=0$

And, for $x=5$,

$x^3-\frac{1}{x^3}=125-\frac{1}{125}=\frac{15624}{125}$