Question

If $y-\frac{1}{y}=9$, find the value of $y^3-\frac{1}{y^3}$

Answer 1

We know that, $x^3-y^3=(x-y{)}^3+3xy(x-y)$
Put $x=y$ and $y=\frac{1}{y},y^3-\frac{1}{y^3}={(y-\frac{1}{y})}^3+3(y-\frac{1}{y})$
$=(9{)}^3+3(9)=729+27=756$