Question 69

If $$x - \frac{1}{x} = 1$$, then what is the value of $$x^8 + \frac{1}{x^8}?$$

Solution

$$x-\frac{1}{x}=1$$

Squaring on both sides,

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

$$x^2+\frac{1}{x^2}=3$$

Squaring on both sides,

$$x^4+\frac{1}{x^4}+2=9$$

$$x^4+\frac{1}{x^4}=7$$

Squaring on both sides,

$$x^8+\frac{1}{x^8}+2=49$$

$$x^8+\frac{1}{x^8}=47$$

Hence, the correct answer is Option C

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