Question

If a/b + b/a = 2
Then find the value of (a/b)^2019 + (b/a)^2019

Answer 1

$\frac{a}{b} + \frac{b}{a} = 2 \\ \frac{ {a}^{2} + {b}^{2} }{ab} = 2 \\ {a}^{2} + {b}^{2} = 2ab \\ {a}^{2} - 2ab + {b}^{2} = 0 \\ {(a - b)}^{2} = 0 \\ a - b = 0 \\ a = b \\ \frac{a}{b} = 1 \\ \frac{b}{a} = 1 \\ \\ now \\ \\ {( \frac{a}{b} )}^{2019} + {( \frac{b}{a}) }^{2019} \\ = {(1)}^{2019} + {(1)}^{2019} \\ = 1 + 1 \\ = 2$