Question

Those who are going to answer these 2 questions by showing steps and explanation I am gonna mark them as brainleist but answer should be correct​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{(\frac{ {x}^{ - b} }{ {x}^{ - a} })^{c} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{a} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{b}=1}}}$

$\green{\tt{\therefore{(\frac{ {x}^{ - b} }{ {x}^{ - a} })^{a + b} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{c + b} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{c + a} =1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{c} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{a} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{b} \\ \\ \tt: \implies (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{(a + b)} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{(c + b)} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{(c + a)}$

$\red{\underline \bold{To \: Find : }} \\ \tt: \implies (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{c} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{a} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{b} = ?\\ \\ \tt: \implies (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{a + b} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{c + b} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{c + a} =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{c} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{a} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{b} \\ \\ \tt: \implies (\frac{ {x}^{a} }{ {x}^{b} })^{c} \times (\frac{ {x}^{b} }{ {x}^{c} })^{a} \times (\frac{ {x}^{c} }{ {x}^{a} })^{b} \\ \\ \tt: \implies \frac{ {x}^{ac} }{ {x}^{bc} } \times \frac{ {x}^{ab} }{ {x}^{ab} } \times \frac{ {x}^{bc} }{{x}^{ac} } \\ \\ \green{\tt: \implies 1} \\ \\ \green{\tt \therefore (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{c} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{a} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{b} = 1} \\ \\ \bold{Similarly : } \\ \tt: \implies (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{a + b} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{c + b} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{c + a}$

$\tt: \implies (\frac{ {x}^{ a} }{ {x}^{b} })^{a + b} \times (\frac{ {x}^{ b} }{ {x}^{c} })^{c + b} \times (\frac{ {x}^{c} }{ {x}^{ a} })^{c + a} \\ \\ \tt: \implies (\frac{ {x}^{ ( {a}^{2} + ab) } }{ {x}^{ (ab+ {b}^{2}) } } \times (\frac{ {x}^{ ({b}^{2} + bc) } }{ {x}^{( {c}^{2} + bc) } }) \times (\frac{ {x}^{( {c}^{2} + ac) } }{ {x}^{ ( {a}^{2} + ac)} })\\ \\ \tt: \implies \frac{ {x}^{ {b}^{2} + bc + {c}^{2} + bc + {c}^{2} + ac} }{ {x}^{ab + {b}^{2} + {c}^{2} + bc + {a}^{2} + ac } } \\ \\ \tt: \implies \frac{ {x}^{ {a}^{2} + {b}^{2} + {c}^{2} + ab +bc + ac } }{ {x}^{ {a}^{2} + {b}^{2} + {c}^{2} + ab + bc + ac } } \\ \\ \green{\tt: \implies 1} \\ \\ \green{\tt \therefore (\frac{ {x}^{ - b} }{ {x}^{ - a} })^{a + b} \times (\frac{ {x}^{ - c} }{ {x}^{ - b} })^{c + b} \times (\frac{ {x}^{ - a} }{ {x}^{ - c} })^{c + a} = 1}$

Answer 2

(i) (x^-b/x^-a)^c × (x^-c/x^-b)^a × (x^-a/x^-c)^b

➫ (x^a-b)^c × (x^b-c)^a × (x^c-a)^b

➫ (x^ac-bc) × (x^ab-ac) × (x^bc-ab)

➫ x^ac-bc+ab-ac+bc-ab

➫ x^ac-ac+ab-ab+bc-bc

➫ x⁰

➫ 1

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

『 $\green{Answer=1}$ 』

(ii) (x^-b/x^-a)^a+b × (x^-c/x^-b)^b+c × (x^-a/x^-c)^c+a

➫ (x^a-b)^a+b × (x^b-c)^b+c × (x^c-a)^c+a

➫ {x^(a-b)(a+b)} × {x^(b-c)(b+c)} × {x^(c-a)(c+a)}

➫ x^(a²-b²) × x^(b²-c²) × x^(c²-a²)

➫ x^a²-b²+b²-c²+c²-b²

➫ x^a²-a²+b²-b²+c²-c²

➫ x⁰

➫ 1

∴ (x^-b/x^-a)^a+b × (x^-c/x^-b)^b+c × (x^-a/x^-c)^c+a = 1

『 $\green{Answer=1}$』