$$p + \frac{1}{{p + 2}} = 1{\text{,}}$$   Find the value of $${\left( {p + 2} \right)^3}$$  $$+$$ $$\frac{1}{{{{\left( {p + 2} \right)}^3}}}$$  $$-$$ 3 = ?

Solution(By Examveda Team)

$$\eqalign{ & {\text{Give,}} \cr & p + \frac{1}{{p + 2}} = 1 \cr & {\text{Adding 2 both sides}} \cr & p + 2 + \frac{1}{{p + 2}} = 1 + 2 = 3 \cr & {\text{Let}}\left( {p + 2} \right) = a\,\,\&\,\, \frac{1}{{p + 2}} = b \cr & a + b = 3 \cr & {\text{Cubbing both sides}} \cr & {\left( {a + b} \right)^3} = {3^3} \cr & {a^3} + {b^3} + 3ab\left( {a + b} \right) = 27 \cr & {a^3} + {b^3} + 3 \times ab \times 3 = 27 \cr & {a^3} + {b^3} + 9ab = 27......\left( {\text{i}} \right) \cr & {\text{Now,}} \cr & a \times b = \left( {p + 2} \right) \times \frac{1}{{\left( {p + 2} \right)}} \cr & a \times b = 1........\left( {{\text{ii}}} \right) \cr & {\text{Put the a & b value in equation}}\left( {\text{i}} \right) \cr & {a^3} + {b^3} + 9 \times 1 = 27 \cr & {a^3} + {b^3} = 27 - 9 = 18 \cr & \therefore {\left( {p + 2} \right)^3} + \frac{1}{{{{\left( {p + 2} \right)}^3}}} - 3 \cr & = {a^3} + {b^3} - 3 \cr & = 18 - 3 \cr & = 15 \cr} $$