If x^4 + 1x^4 = 25716, then find 813 ( x^3 + 1x^3 ), where x > 0.

Answer: Option A

$$\begin{array}{rl}& x^4+\frac{1}{x^4}=\frac{257}{16}\\ & {\left(x^2+\frac{1}{x^2}\right)}^2=\frac{257}{16}+2\\ & {\left(x^2+\frac{1}{x^2}\right)}^2={\left(\frac{17}{4}\right)}^2\\ & x^2+\frac{1}{x^2}=\frac{17}{4}\\ & x+\frac{1}{x}={\left(\frac{17}{4}+2\right)}^{\frac{1}{2}}=\frac{5}{2}\\ & \text{Here }x=2\\ & \text{Hence }\frac{8}{13}\left(x^3+\frac{1}{x^3}\right)\\ & =\frac{8}{13}\left(8+\frac{1}{8}\right)\\ & =\frac{8}{13}\times \frac{65}{8}\\ & =5\end{array}$$