Should I Raise It To The Fifth Power?

Relevant wiki: De Moivre's Theorem - Raising to a Power - Intermediate

Raising both sides to the power of 5 can be extremely tedious and requires extensive work, however we can use de Moivre's theorem to greatly simplify the working:

I am using Newton's sums or identities to solve the problem.

Let $P_n = x^n + \dfrac 1{x^n}$ , where $n$ is a positive integer, $S_1 = x + \dfrac 1x = 2\cos 42^\circ$ and $S_2 = x \times \dfrac 1x = 1$ . Then we have:

$\begin{aligned} P_1 & = S_1 = 2\cos 42^\circ & \implies x + \dfrac 1x = 2\cos 42^\circ \end{aligned}$

$\begin{aligned} P_2 & = S_1P_1 - 2S_2 \\ & = 4\cos^2 42^\circ - 2 \\ & = 2\cos 84^\circ & \implies x^2 + \dfrac 1{x^2} = 2\cos 84^\circ \end{aligned}$

$\begin{aligned} P_3 & = S_1P_2 - S_2P_1 \\ & = (2\cos 42^\circ)(4\cos^2 42^\circ - 2) - 2\cos 42^\circ \\ & = 2(4\cos^3 42^\circ - 3 \cos 42^\circ) \\ & = 2\cos 126^\circ \end{aligned}$

$\begin{aligned} \implies x^3 + \dfrac 1{x^3} & = 2\cos 126^\circ \end{aligned}$

$\begin{aligned} P_4 & = S_1P_3 - S_2P_2 \\ & = 16\cos^4 42^\circ - 12\cos^2 42^\circ - 4\cos^2 42^\circ + 2 \\ & = 2(8\cos^4 42^\circ - 8\cos^2 42^\circ + 1) \\ & = 2\cos 168^\circ \end{aligned}$

$\begin{aligned} \implies x^4 + \dfrac 1{x^4} & = 2\cos 168^\circ \end{aligned}$

$\begin{aligned} P_5 & = S_1P_4 - S_2P_3 \\ & = 32\cos^5 42^\circ - 32\cos^3 42^\circ + 4\cos 42^\circ - 8\cos^3 42^\circ + 6 \cos 42^\circ \\ & = 2(16\cos^5 42^\circ - 20\cos^3 42^\circ + 5\cos 42^\circ) \\ & = 2\cos 210^\circ \end{aligned}$

$\begin{aligned} \implies x^5 + \dfrac 1{x^5} & = 2\cos 210^\circ = 2 \left(-\frac {\sqrt 3}2 \right) = -\sqrt 3 \end{aligned}$

$\implies A = \boxed{3}$ .

Here is what I did : $x + \frac{1}{x} = 2\cos42^\circ$ Squaring both sides, we have, $x^{2}+\frac{1}{x^{2}} = 4\cos^{2}42^\circ - 2= 2\cos82^\circ \ldots 1$ Now cube the first equation to give, $x^{3} + \frac{1}{x^{3}} + 3(x+ \frac{1}{x}) = 8 \cos^{3}42^\circ$ Using the value of first or initial equation and identity, $4\cos^{3}\theta = \cos3\theta + 3\cos\theta$ $\Rightarrow x^{3} + \frac{1}{x^{3}} = 2\cos126^\circ \ldots 2$ Now multiply 1 and 2 , To get, $x^{5} + \frac{1}{x^{5}} + x + \frac{1}{x} = 2 (\cos210^\circ + \cos42^\circ)$ $x^{5} + \frac{1}{x^{5}} = -\sqrt {3}$ So, $A = \boxed {3}$

Given $x+\dfrac 1x = 2\cos 42^\circ$

Let $\theta = 42^\circ$ . Cross - Multiplying and Rearranging, $x^2-2\cos \theta \cdot x+ 1=0\\x = \cos\theta \pm \sqrt{\cos^2\theta -1}=cos\theta \pm i\sin\theta=e^{\pm i\theta}$

So, $x^5 = e^{\pm 5i\theta}\\x^5+\dfrac 1{x^5} = 2~\Re\{ x^5\} = 2\cos \left(\pm 5\theta\right) = 2\cos (180^\circ + 30^\circ) = -\sqrt3$

Therefore, $\color{#3D99F6}{A=\boxed{ 3}}$