Function can be tricky (3)

Since $f \left(x + \dfrac 1x\right) = x^5 + \dfrac 1{x^5}$ , then if $a + \dfrac 1a = 3$ $\implies f(3) = a^5 + \dfrac 1{a^5}$ .

The problem can be solved using Newton's sums method . Let $P_n = a^n + \dfrac 1{a^n}$ , $S_1 = a + \dfrac 1a = 3$ and $S_2 = a \cdot \dfrac 1a = 1$ . Then we have:

\(\begin{array} {} P_1 = S_1 & \implies a + \dfrac 1a = 3 & = 3 \\ P_2 = S_1P_1 - 2S_2 & \implies a^2 + \dfrac 1{a^2} = 3\cdot 3 - 2 & = 7 \\ P_3 = S_1P_2 - S_2P_1 & \implies a^3 + \dfrac 1{a^3} = 3\cdot 7 - 1\cdot 3 & = 18 \\ P_4 = S_1P_3 - S_2P_2 & \implies a^4 + \dfrac 1{a^4} = 3\cdot 18 - 1\cdot 7 & = 47 \\ P_5 = S_1P_4 - S_2P_3 & \implies a^5 + \dfrac 1{a^5} = 3\cdot 47 - 1\cdot 18 & = 123 \end{array} \)

$\implies f(3) = \boxed{123}$

Let $y$ = $x+\frac{1}{x}$ :

$(x+\frac{1}{x})^2 = x^2+2(x)(\frac{1}{x})+\frac{1}{x^2}$

$x^2+\frac{1}{x^2} = y^2-2$

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$x^3+\frac{1}{x^3} = (x+\frac{1}{x})(x^2-(x)(\frac{1}{x})+\frac{1}{x^2})$

$x^3+\frac{1}{x^3} = (y)(y^2-2-1)$

$x^3+\frac{1}{x^3} = y^3-3y$

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$(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3}) = x^5+\frac{1}{x^5}+x+\frac{1}{x}$

$x^5+\frac{1}{x^5} = (y^2-2)(y^3-3y)-y$

$x^5+\frac{1}{x^5} = y^5-3y^3-2y^3+6y-y$

$x^5+\frac{1}{x^5} = y^5-5y^3+5y$

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$\Rightarrow f(y)=y^5-5y^3+5y$

$f(3)=3^5-5(3)^3+5(3)$

$f(3)=123$

Here's a solution reverse engineered from @TommyLi 's solution using simple polynomial division : Solution

$y=\frac{x^2+1}{x} \ \implies f(x+\frac{1}{x})=f(y)$

$f(\frac{x^2+1}{x})=x^5+\frac{1}{x^5}=\frac{x^{10}+1}{x^5}=\left(\frac{x^2+1}{x}\right)\left(\frac{x^8-x^6+x^4-x^2+1}{x^4}\right)=$

$=\left(\frac{x^2+1}{x}\right)\left(\left(\frac{x^2+1}{x}\right)^4-\frac{5x^6+5x^4+5x^2}{x^4}\right)=$

$=\left(\frac{x^2+1}{x}\right)\left(\left(\frac{x^2+1}{x}\right)^4-5\left(\frac{x^2+1}{x}\right)^2+5\frac{x^2}{x^2}\right)=$

$=y(y^4-5y^2+5)$

$y=3 \implies f(3)=3(3^4-5\cdot3^2+5)=123$

(x+1/x)^5=(x^5+1/x^5)+5(x+1/x)^3-5(x+1/x) (Using binomial theorem) So,f(x+1/x)=(x+1/x)^5 - 5(x+1/x)^3 +5(x+1/x) Put x+1/x=3 So f(3)=243-135+15 =123