If $(x + \frac{1}{x}) = \sqrt{13},$ $\left( {x + \frac{1}{x}} \right){\text{ = }}\sqrt {13} {\text{,}}$ then the value of $(x^{3} + \frac{1}{x^{3}})$ $\left( {{x^3} + \frac{1}{{{x^3}}}} \right)$ is = ?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (x + \frac{1}{x}) = \sqrt{13} \\ \Rightarrow {(x + \frac{1}{x})}^{2} - 4 = {(\sqrt{13})}^{2} \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 9 \\ \Rightarrow (x + \frac{1}{x}) = 3 \\ \Rightarrow {(x + \frac{1}{x})}^{3} = 3^{3} = 27 \\ \Rightarrow x^{3} - \frac{1}{x^{3}} - 3. x . \frac{1}{x} (x - \frac{1}{x}) = 27 \\ \Rightarrow x^{3} - \frac{1}{x^{3}} - 3 \times 3 = 27 \\ \Rightarrow x^{3} - \frac{1}{x^{3}} = 27 + 9 = 36 \end{aligned}

$\eqalign{ & \left( {x + \frac{1}{x}} \right) = \sqrt {13} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} - 4 = {\left( {\sqrt {13} } \right)^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow \left( {x + \frac{1}{x}} \right) = 3 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {3^3} = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.x.\frac{1}{x}\left( {x - \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times 3 = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 27 + 9 = 36 \cr}$