If \(x^4+\displaystyle\frac{1}{x^4}=6239\) and \(x>0\), then find the value of \(x^3+\displaystyle\frac{1}{x^3}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 702
Work back with \(\left(t+\frac{1}{t}\right)^2=t^2+\frac{1}{t^2}+2\) to reach \(x+\frac{1}{x}\); then the cube follows from \(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\).
\(\left(x^2+\displaystyle\frac{1}{x^2}\right)^2=6239+2=6241\), so \(x^2+\displaystyle\frac{1}{x^2}=\sqrt{6241}=79\)
\(\left(x+\displaystyle\frac{1}{x}\right)^2=79+2=81\), and \(x>0\), so \(x+\displaystyle\frac{1}{x}=9\)
\(x^3+\displaystyle\frac{1}{x^3}=9^3-3\times9=729-27=702\)