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

Answer:- 1

Explanation:-

Solution:

\begin{aligned} p + \frac{1}{p + 2} = 1, Find the value of {(p + 2)}^{3} + \frac{1}{{(p + 2)}^{3}} - 3 \\ A d d i n g 2 o n both sides \\ \Rightarrow (p + 2) + \frac{1}{(p + 2)} - 3 \\ Taking cube on both sides \\ \Rightarrow {(p + 2)}^{3} + \frac{1}{{(p + 2)}^{3}} + 3 (p + 2) \times \frac{1}{(p + 2)} \\ \Rightarrow [(p + 2) + \frac{1}{(p + 2)}] = {[(p + 2) + \frac{1}{(p + 2)}]}^{3} \\ \Rightarrow [(p + 2) + \frac{1}{(p + 2)}] = 3^{3} \\ \Rightarrow [(p + 2) + \frac{1}{(p + 2)}] = 27 \\ \Rightarrow {(p + 2)}^{3} + \frac{1}{{(p + 2)}^{3}} + 9 = 27 \\ \Rightarrow {(p + 2)}^{3} + \frac{1}{{(p + 2)}^{3}} = 18 \\ According to the question, \\ 18 - 3 = 15 \end{aligned}

$\eqalign{ & p + \frac{1}{{p + 2}} = 1{\text{, Find the value of }}{\left( {p + 2} \right)^3} + \frac{1}{{{{\left( {p + 2} \right)}^3}}} - 3 \cr & Adding2on{\text{ both sides}} \cr & \Rightarrow \left( {p + 2} \right) + \frac{1}{{\left( {p + 2} \right)}} - 3 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {p + 2} \right)^3} + \frac{1}{{{{\left( {p + 2} \right)}^3}}} + 3\left( {p + 2} \right) \times \frac{1}{{\left( {p + 2} \right)}} \cr & \Rightarrow \left[ {\left( {p + 2} \right) + \frac{1}{{\left( {p + 2} \right)}}} \right] = {\left[ {\left( {p + 2} \right) + \frac{1}{{\left( {p + 2} \right)}}} \right]^3} \cr & \Rightarrow \left[ {\left( {p + 2} \right) + \frac{1}{{\left( {p + 2} \right)}}} \right] = {3^3} \cr & \Rightarrow \left[ {\left( {p + 2} \right) + \frac{1}{{\left( {p + 2} \right)}}} \right] = 27 \cr & \Rightarrow {\left( {p + 2} \right)^3} + \frac{1}{{{{\left( {p + 2} \right)}^3}}} + 9 = 27 \cr & \Rightarrow {\left( {p + 2} \right)^3} + \frac{1}{{{{\left( {p + 2} \right)}^3}}} = 18 \cr & {\text{According to the question,}} \cr & 18 - 3 = 15 \cr}$