Question

If $x+\frac{1}{x}=1$ then $x^{2018}+\frac{1}{x^{2018}}$ is equal to

• A. $0$
• B. $-1$
• C. $1$
• D. $2018$

Answer 1

The correct option is B $-1$

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

Multiplying through by $x$ to get

$\Rightarrow$ $x^2+1=x$

Subtract $x$ from both sides to get

$\Rightarrow$ $x^2-x+1=0$

Multiplying by $x+1$ this becomes,

$\Rightarrow$ $(x+1)(x^2-x+1)=0$

$\Rightarrow$ $x^3+1=0$

$\Rightarrow$ $x^3=-1$

Now,

$\Rightarrow$ $x^{2018}+\frac{1}{x^{2018}}=\frac{x^{2019}}{x}+\frac{x}{x^{2019}}$

$=\frac{(x^3{)}^{673}}{x}+\frac{x}{(x^3{)}^{673}}$

$=\frac{(-1{)}^{673}}{x}+\frac{x}{(-1{)}^{673}}$

$=-(\frac{1}{x}+x)$

$=-1$