Question

If $x=3+2\sqrt{2}$ find the value of $x^4+\frac{1}{x^4}$

Answer 1

$\begin{array}{c}x=3+2\sqrt{2}\\ x^4+\frac{1}{x^4}=?\\ x={\left(\sqrt{2}+1\right)}^2\\ {\left(\sqrt{2}+1\right)}^8+{\left(\sqrt{2}-1\right)}^8\\ \left[8C_0{\left(\sqrt{2}\right)}^8+8C_1{\left(\sqrt{2}\right)}^7+8C_2{\left(\sqrt{2}\right)}^6+...........\right]\\ =2\left[8C_0{\left(\sqrt{2}\right)}^8+8C_2{\left(\sqrt{2}\right)}^6+8C_4{\left(\sqrt{2}\right)}^4+8C_6{\left(\sqrt{2}\right)}^2+8C_8\right]\\ =2\left[16+28\times 8+70\times 4+28\times 2+1\right]\\ =2\left[16+280+280+1\right]=\left(560+17\right)\times 2\\ =577\times 2\\ =1154\end{array}$

Hence,

Solved.