Question

If $x=3+\sqrt{8}$, then find the value of $x^4+\frac{1}{x^4}$

Answer 1

Given $x=3+\sqrt{8}$

So $\frac{1}{x}=\frac{1}{3+\sqrt{8}}\cdot \frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}=3-\sqrt{8}$

Thus $x+\frac{1}{x}=3+\sqrt{8}+3-\sqrt{8}=6$

$\Rightarrow x^2+\frac{1}{x^2}+2\cdot x\cdot \frac{1}{x}=6^2=36$, square both sides

$\Rightarrow x^2+\frac{1}{x^2}=36-2=34$

$\Rightarrow x^4+\frac{1}{x^4}+2\cdot x^2\cdot \frac{1}{x^2}={34}^2$, square both sides

$\therefore x^4+\frac{1}{x^4}={34}^2-2=1154$