Algebraic Operations

Question

How to find the value of $E=x^5 + x^{-5}$?. Knowing that $x^3 + x^{-3}=54$. At first it seemed like a simple exercise, but when I started to solve it I had difficulties. Try the "Remarkable Products", but I did not find it. Try to express $E$ in terms of $x^3 + x^{-3}$ and then replace, but I did not either. Is there something that is forgetting me?, Is there any way to solve this exercise ?.

Answer 1

One way is to solve the equivalent "quadratic" equation

$$x^6-54x^3+1=0$$

Note $x$ must have a positive value, and due to the symmetry of your equation it wont matter which of the two you take for $x$ (the positive solutions are reciprocals of one another).

Answer 2

If $x^3+x^{-3}=a$ then the binomial theorem implies $$\begin{align}x^9+3x^3+3x^{-3}+x^{-9}&=a^3\\ x^{15}+5x^{9}+10x^3+10x^{-3}+5x^{-9}+x^{-15}&=a^5 \end{align}$$ so that $$\begin{align}x^9+x^{-9}&=a^3-3a\\ x^{15}+x^{-15}&=a^5-5a^3+5a\text{.} \end{align}$$

Likewise, if $x^5+x^{-5}=E$ then $$x^{15}+3x^5+3x^{-5}+x^{-15}=E^3$$ so that $$x^{15}+x^{-15}=E^3-3E\text{.}$$

Therefore $$E^3-3E=a^5-5a^3+5a\text{,}$$ a cubic equation for $E$.

Answer 3

$$x^3 + x^{-3} = 54 \implies x^6 - 54x^3 + 1 = 0$$ Let $y = x^3$. The equation above becomes $$y^2 - 54y + 1 = 0$$ Solve the quadratic equation to find roots for $y$. Then find values for $x$ using $x = \sqrt[3]{y}$. Note that $x$ has $6$ roots. Once you have these, finding $E$ is trivial.