High Phower

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If $(x+1)(x-1)=x$ , evaluate $x^{10}+x^{-10}$ .

The answer is 123.

1 solution

Pi Han Goh
Dec 20, 2013

$\begin{aligned} x^2 - 1 & = & x \\ x - \frac {1}{x} & = & 1 \\ \left ( x - \frac {1}{x} \right )^5 & = & 1 \\ \left ( x^5 - \frac {1}{x^5} \right ) - 5 \left ( x^3 - \frac {1}{x^3} \right ) + 10 \left ( x - \frac {1}{x} \right ) & = & 1 \\ \left ( x^5 - \frac {1}{x^5} \right ) - 5 \left ( \left (x - \frac {1}{x} \right )^3 - 3 \left (x - \frac {1}{x} \right ) \right ) + 10 (1) & = & 1 \\ \left ( x^5 - \frac {1}{x^5} \right ) - 5 \left ( 1^3 + 3(1) \right ) + 10 (1) & = & 1 \\ x^5 - \frac {1}{x^5} & = & 11 \\ \left ( x^5 - \frac {1}{x^5} \right )^2 & = & 11^2 \\ \left ( x^{10} + \frac {1}{x^{10}} \right ) - 2 & = & 121 \\ x^{10} + \frac {1}{x^{10}} & = & \boxed{123} \\ \end{aligned}$

here's the solution: first observe:

if $x+\frac{1}{x}=p$ then the desired expression is equal to :

$(((p^2-2)^2-2)(p)-((p)^3-3p))^2-2$ .Observe

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

since $(x+1)(x-1)=x$ it follows that $x-\frac{1}{x}=1$ . where we obtain $x+\frac{1}{x}=\sqrt{5}$ substituting this value for $p$ we obtain :

$x^{10}+\frac{1}{x^{10}}=\boxed{123}$

Lorenc Bushi - 7 years, 5 months ago

Genious! Bravo, bravo!

Guilherme Dela Corte - 7 years, 5 months ago

High Phower

The answer is 123.

1 solution

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