Question

solve
(x^a/x^-b)^a-b × (x^b/x^-c)^b-c × (x^c/x^-a)^c-a =1​

Answer 1

TO PROVE :–

$\\ \bf \implies \left( \dfrac{x^a}{x^{-b}} \right)^{a-b} \times \left( \dfrac{x^b}{x^{-c}} \right)^{b-c}\times \left( \dfrac{x^c}{x^{-a}} \right)^{c-a} =1$

SOLUTION :–

• Let's take L.H.S. –

$\\ \bf \: \: = \: \: \left( \dfrac{x^a}{x^{-b}} \right)^{a-b} \times \left( \dfrac{x^b}{x^{-c}} \right)^{b-c}\times \left( \dfrac{x^c}{x^{-a}} \right)^{c-a}$

$\\ \bf \: \: = \: \:(x^{a + b} )^{a-b} \times (x^{b + c})^{b-c} \times (x^{c + a} )^{c - a}$

• We know that –

$\\ \bf \longrightarrow \pink{{({a}^{m})}^{n} = {a}^{mn}}$

• So that –

$\\ \bf \: \: = \: \:(x)^{(a + b)(a-b)}\times(x)^{(b + c)(b-c)} \times (x)^{(c + a)(c - a)}$

$\\ \bf \: \: = \: \:(x)^{ {a}^{2} - {b}^{2} }\times(x)^{ {b}^{2} - {c}^{2} } \times (x)^{ {c}^{2} - {a}^{2}}$

$\\ \bf \: \: = \: \:(x)^{({a}^{2} - {b}^{2}) + ({b}^{2} - {c}^{2}) + ({c}^{2} - {a}^{2})}$

$\\ \bf \: \: = \: \:(x)^{({a}^{2} - {b}^{2}+{b}^{2} - {c}^{2}+{c}^{2} - {a}^{2})}$

$\\ \bf \: \: = \: \:(x)^{0}$

$\\ \bf \: \: = \: \:1$

$\\ \bf \: \: = \: \:R.H.S.$

$\\ \bf \implies Hence \: \: Proved$

Answer 2