An algebra problem by yaman malla

$\displaystyle \color{#D61F06}{x + \dfrac 1x = 3}$

$\displaystyle (x + \dfrac 1x)^2 = 9$

$x^2 + \dfrac {1}{x^2} + 2 = 9 \Longrightarrow \color{#CEBB00}{x^2 + \dfrac {1}{x^2} = 7}$

$\displaystyle (x + \dfrac 1x)(x^2 + \dfrac {1}{x^2}) = {\color{#D61F06}{3}} \times {\color{#CEBB00}{7}} = 21$

$x^3 + \dfrac {1}{x^3} + x + \dfrac 1x = 21$

$\displaystyle {\color{#20A900}{x^3 + \dfrac {1}{x^3}}} = 21 - (x + \dfrac 1x) = 21 - {\color{#D61F06}{3}} = \color{#20A900}{18}$

$\displaystyle (x^2 + \dfrac {1}{x^2})^2 = 49$

$x^4 + \dfrac {1}{x^4} + 2 = 49 \Longrightarrow \color{#EC7300}{x^4 + \dfrac {1}{x^4} = 47}$

$\displaystyle (x + \dfrac 1x)(x^4 + \dfrac {1}{x^4}) = {\color{#D61F06}{3}} \times {\color{#EC7300}{47}}$

$x^5 + \dfrac {1}{x^5} + x^3 + \dfrac {1}{x^3} = 141$

$x^5 + \dfrac {1}{x^5} = 141 - (x^3 + \dfrac {1}{x^3}) = 141 - \color{#20A900}{18}$

$\color{#3D99F6}{x^5 + \dfrac {1}{x^5}} = 123$

The fun solution

Let $L_n$ be the Lucas sequence, defined recursively by $L_0 = 2, L_1 = 1, L_n = L_{n-1} + L_{n-2}\ \text{for}\ n \geq 2.$ (These are like the Fibonacci numbers, but with different initial values.)

Then $L_n = \phi^n + \bar\phi^n$ , where $\phi = \tfrac12+\tfrac12\sqrt 5$ and $\bar\phi = \tfrac12-\tfrac12\sqrt 5 = -1/\phi$ .

Now $x + 1/x = 3 = L_2$ implies that $\{x,\tfrac1x\} = \{\phi^2,\bar\phi^2\}$ .

Immediately we find that $x^5 + \frac1{x^5} = (\phi^2)^5 + (\bar\phi^2)^5 = \phi^{10} + \bar\phi^{10} = L_{10}.$ Generating the tenth Lucas number is not hard: $L_{0\dots} = (2,1,3,4,7,11,18,29,47,76,\boxed{123},\dots).$

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The boring solution

Let $N = x + 1/x$ .

To generate something with $x^5$ and $1/x^5$ in it, I start with the fifth power of the given expression: $N^5 = \left(x+\frac1x\right)^5 = x^5 + 5x^3 + 10x + \cdots$ (The dots represent the same terms with $1/x$ instead of $x$ .) This is a good start, but we need to get rid of the $x^3$ term. Therefore consider $N^3 = \left(x+\frac1x\right)^3 = x^3 + 3x + \cdots$ So we'll subtract $N^5 - 5N^3 = (x^5 + 5x^3 + 10x + \cdots) - 5(x^3 + 3x + \cdots) = x^5 - 5x + \cdots$ Finally, add $5N$ to get $N^5 - 5N^3 + 5N = x^5 + \frac{1}{x^5}.$ Since $N = x + 1/x = 3$ , we find the solution $N^5 - 5N^3 + 5N = 3^5 - 5\cdot 3^3 + 5\cdot 3 = \boxed{123}.$

Take both sides of $3 = x + \frac{1}{x}$ to the third and fifth powers, grouping terms in each case: $27 = \left(x + \frac{1}{x}\right)^3 = \left(x^3 + \frac{1}{x^3}\right) + 3\left(x + \frac{1}{x}\right)$ $243 = \left(x + \frac{1}{x}\right)^5 = \left(x^5 + \frac{1}{x^5}\right) + 5\left(x^3+\frac{1}{x^3}\right) + 10\left(x+\frac{1}{x}\right)$

Solving for $x^3+\frac{1}{x^3}$ in the first equation: $x^3 + \frac{1}{x^3} = 27 - 3\left(x + \frac{1}{x}\right) = 27 - 3(3) = 18$ and then solving for $x^5 + \frac{1}{x^5}$ in the second equation: $x^5 + \frac{1}{x^5} = 243 - 5\left(x^3+\frac{1}{x^3}\right) - 10\left(x+\frac{1}{x}\right) = 243 - 5(18) - 10(3) = \boxed{123}$

Using Binomial expansion $\left(x+\dfrac{1}{x}\right)^5 = \binom{5}{0}x^5 + \binom{5}{1}x^4\cdot\dfrac{1}{x} + \binom{5}{2}x^3\cdot\dfrac{1}{x^2}+\binom{5}{3}x^2\cdot\dfrac{1}{x^3} + \binom{5}{4}x\cdot\dfrac{1}{x^4}+\binom{5}{5}\cdot\dfrac{1}{x^5} \\ \left(x+\dfrac{1}{x}\right)^5 = x^5+5x^3 +10x + \dfrac{10}{x} + \dfrac{5}{x^3} +\dfrac{1}{x^5} \\ 243 = \left(x^5+\dfrac{1}{x^5}\right) +10\left(x+ \dfrac{1}{x}\right) +5\underbrace{{\color{#3D99F6}\left(x^3+\dfrac{1}{x^3}\right)}}_{\text{Note}} \implies \left(x^5 +\dfrac{1}{x^5}\right) = 243 -30 - 90 =\boxed{ 123 }$

Note:

$\left(x+\dfrac{1}{x}\right)^3 = x^3 + \dfrac{1}{x^3} +\dfrac{3x^2}{x} +\dfrac{3x}{x^2}=x^3+\dfrac{1}{x^3} +3\left(x+\dfrac{1}{x}\right) \\x^3+\dfrac{1}{x^3} = 3^3 -3\,(3) = 18$

First square and cube both sides to the equation to get

$x^2 + \dfrac{1}{x^2} = 7$

and

$x^3 + \dfrac{1}{x^3} = 18$

Multiplying these two equation gives

$x^5 + \dfrac{1}{x^5} + \underbrace{\left(x + \dfrac 1x \right)}_{= \ 3} = 126 \\ \therefore \ x^5 + \dfrac{1}{x^5} = \boxed{123}$

Let $f_n(x)=x^n+x^{-n}$ And given that $f_1(x)=3$ Multiplying this on both sides, we get $3f_n(x)=x^{n+1}+x^{n-1}+x^{-n+1}+x^{-n-1}=f_{n+1}(x)+f_{n-1}(x)$ Thus we have obtained a recurrence relation where $f_0(x)=2$ (trivially) and $f_1(x)=3$ and $f_{n+2}(x)=3f_{n+1}(x)-f_n(x)$

Since we need to only find $f_5(x)$ we need not solve this reccurence but instead just get next values step wise. However, if a significantly large number is to be calculated like $f_{2017}(x)$ then It would be much easier to solve the recurrence relation and plug the values.

Using the former method, $f_2(x)=3*3-2=7$ $f_3(x)=3*7-3=18$ $f_4(x)=3*18-7=47$ $f_5(x)=3*47-18=123$ and using the latter equation, $f_n(x) = 2^{-n} ((3 - \sqrt{5})^n + (3 + \sqrt{5})^n)$ or $f_n(x)≈0.381966^n + 2.61803^n$ $f_5(x)=122.990+0.00813=122.998$

Rather than raising the former expression to the fifth power, we can solve for $x$ in the former expression using the quadratic formula and then plug the answer into the latter expression. First, multiply all terms by $x$ and rearrange. $x^{2} + 1 = 3x$ $x^{2} - 3x + 1 = 0$ Next, use the quadratic formula ( $x = \frac{-b \pm\sqrt{b^{2}-4ac}}{2a}$ ) to solve for $x$ . For clarity's sake, $a = 1$ , $b = -3$ , $c = 1$ . $x = \frac{3 \pm\sqrt{(-3)^{2}-4}}{2}$ Simplify, $x = \frac{3 \pm\sqrt{5}}{2}$ Both $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ are valid solutions; however, I will only use $\frac{3+\sqrt{5}}{2}$ for simplicity's sake. Plugging this answer into the latter expression, we get ${\left(\frac{3+\sqrt{5}}{2}\right)}^5 +{\left(\frac{1}{\frac{3+\sqrt{5}}{2}}\right)}^5 \Longrightarrow {\left(\frac{3+\sqrt{5}}{2}\right)}^5 + {\left(\frac{2}{3+\sqrt{5}}\right)}^5 = \boxed{123}$

We define $F_n = x^n + 1/x^n$ and there exists a recursion for this sequence: $F_n = F_1F_{n-1} - F_{n-2} = 3F_{n-1} - F_{n-2}.$

Thus, we can calculate $F_1 = 3, F_2 = 7, F_3 = 18, F_4 = 47, F_5 = 123.$

Appylying Newton's binomail theorem for (x+ $\frac{1}{x}$ )^5 we will obatin: x^5+5 x^3+10 x+ $\frac{10}{x}$ + $\frac{5}{x^3}$ + $\frac{1}{x^5}$ =3^5=243 Now we will group the therms in a convenable way. x^5+ $\frac{1}{x^5}$ +5(x^3+ $\frac{1}{x^3}$ )+10(x+ $\frac{1}{x}$ )=243 but we know that x+ $\frac{1}{x}$ =3, so . x^5+ $\frac{1}{x^5}$ +5(x^3+ $\frac{1}{x^3}$ )=243-10 3=213. Now we apply Newton's binomail theorem for (x+ $\frac{1}{x}$ )^3=3^3=27. I'll skip to grouping. x^3+ $\frac{1}{x^3}$ +3(x+ $\frac{1}{x}$ )=27 => x^3+ $\frac{1}{x^3}$ =27-8=18. x^5+ $\frac{1}{x^5}$ +5 18=213, so x^5+ $\frac{1}{x^5}$ =213-(5*18)=123. No need to apply the binome for x∈[1;5] ∩N.

x+1/x=3, so x²+1=3x, x^5=x³(3x-1)=3x⁴-x³=(3x-1)(3x²-x)= (3x-1)(3(3x-1)-x)=(3x-1)(8x-3)= 24(3x-1)-17x+3=55x-21 so x^5+1/x^5=((55x-21)²+1)/(55x-21) =(3025(3x-1)-2310x+442)/(55x-21)= (6765x-2583)/(55x-21)=123