If (p – x)(p – y) = 1 and $(x-y=\sqrt{5})$ then find the value of $(p-y{)}^3-\frac{1}{(p-y{)}^3}$.

$(x-y{)}^3=x^3-y^3-3xy(x-y)$

$\Rightarrow (p-x)(p-y)=1$

$\Rightarrow (p-x)=\frac{1}{(p-y)}$ ...(i), $x-y=\sqrt{5}$ ...(ii)

$\Rightarrow (p-y)-(p-x)=\sqrt{5}$ ...(iii)

$\Rightarrow (p-y)-\frac{1}{(p-y)}=\sqrt{5}$(taking cubes on both sides)

$\Rightarrow (p-y{)}^3-\frac{1}{(p-y{)}^3}-3(\sqrt{5})=5\sqrt{5}$
$\Rightarrow (p-y{)}^3-\frac{1}{(p-y{)}^3}=8\sqrt{5}$