Practicing Success
If $x-\frac{1}{x}=1$, then what is the value of $x^8+\frac{1}{x^8} ?$ |
3 119 47 -1 |
47 |
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 |