If \(x^2+\displaystyle\frac{1}{x^2}=23\) and \(x>0\), then find the value of \(x^5+\displaystyle\frac{1}{x^5}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2525
The fifth power comes from multiplying the second and third powers: \(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x^5}+x+\frac{1}{x}\). So \(x+\frac{1}{x}\) and \(x^3+\frac{1}{x^3}\) are found first.
\(\left(x+\displaystyle\frac{1}{x}\right)^2=23+2=25\); since \(x>0\), \(x+\displaystyle\frac{1}{x}=5\)
\(x^3+\displaystyle\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=125-15=110\)
\(23\times110=x^5+\displaystyle\frac{1}{x^5}+5\)
\(x^5+\displaystyle\frac{1}{x^5}=2530-5=2525\)