If $x-\frac{1}{x}=1$, then what is the value of $x^8+\frac{1}{x^8} ?$
Answer & explanation
Correct answer: option 3
If $K+\frac{1}{K}=n$
then, $K^2+\frac{1}{K^2}$ = n2 – 2
If $K-\frac{1}{K}=n$
then, $K^2+\frac{1}{K^2}$ = n2 + 2
If $x-\frac{1}{x}=1$
then, $x^2+\frac{1}{x^2}$ = 12 + 2 = 3
$x^4+\frac{1}{x^4}$ = 32 – 2 = 7
And the value of $x^8+\frac{1}{x^8} $ = 72 – 2 = 47