Question

$\frac{a}{b} + \frac{b}{a}=2 then (\frac{a}{b})^100 - (\frac{b}{a})^100 =$

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:\dfrac{a}{b} + \dfrac{b}{a} = 2$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: {\bigg(\dfrac{a}{b} \bigg) }^{100} - {\bigg(\dfrac{b}{a} \bigg) }^{100}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{a}{b} + \dfrac{b}{a} = 2$

$\rm :\longmapsto\:\dfrac{ {a}^{2} + {b}^{2} }{ab}= 2$

$\rm :\longmapsto\:{a}^{2} + {b}^{2} = 2ab$

$\rm :\longmapsto\:{a}^{2} + {b}^{2} - 2ab = 0$

$\rm :\longmapsto\: {(a - b)}^{2} = 0$

$\rm :\longmapsto\: a - b= 0$

$\bf\implies \:a = b$

Now, Consider,

$\rm :\longmapsto\: {\bigg(\dfrac{a}{b} \bigg) }^{100} - {\bigg(\dfrac{b}{a} \bigg) }^{100}$

On substituting a = b, we get

$\rm \:  =  \:  \: \: {\bigg(\dfrac{b}{b} \bigg) }^{100} - {\bigg(\dfrac{b}{b} \bigg) }^{100}$

$\rm \:  =  \:  \: {1}^{100} - {1}^{100}$

$\rm \:  =  \:  \: 1 - 1$

$\rm \:  =  \:  \: 0$

Hence,

$\bf :\longmapsto\: {\bigg(\dfrac{a}{b} \bigg) }^{100} - {\bigg(\dfrac{b}{a} \bigg) }^{100} = 0$

Additional Information :-

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)