Question

please answer earlier​

Answer 1

Step-by-step explanation:

this is proved that left hand side = right hand side

Answer 2

Given :

$\rm \longrightarrow {\bigg( {x}^{ \bf\frac{b + c}{c - a } } \bigg)}^{ \frac{1}{a - b\: } } {\rm \bigg( {x}^{ \bf\frac{ c + a}{a - b} } \bigg)}^{ \frac{1}{b - c} } \: {\rm \bigg( {x}^{ \bf\frac{ a + b}{b - c} } \bigg)}^{ \frac{1}{c - a} } = 1$

To Prove :

LHS = RHS

Proof :

RHS = 1

$\rm \longrightarrow LHS = {\bigg( {x}^{ \bf\frac{b + c}{c - a } } \bigg)}^{ \frac{1}{a - b\: } } {\rm \bigg( {x}^{ \bf\frac{ c + a}{a - b} } \bigg)}^{ \frac{1}{b - c} } \: {\rm \bigg( {x}^{ \bf\frac{ a + b}{b - c} } \bigg)}^{ \frac{1}{c - a} }$

$\rm \longrightarrow \bigg( {x}^{ \bf\frac{b + c}{(c - a)(a - b)} } \bigg) \rm \bigg( {x}^{ \bf\frac{ c + a}{(a - b)( b - c)} } \bigg)\rm \bigg( {x}^{ \bf\frac{ a + b}{(c - a)( b - c)} } \bigg) \\ \\ \\ \large\rm \longrightarrow {x}^{ \bf\frac{b + c}{(c - a)(a - b)} + \bf\frac{ c + a}{(a - b)( b - c)} + \bf\frac{ a + b}{(c - a)( b - c)}} \\ \\ \\ \rm \large \longrightarrow {x}^{ \bf\frac{(b + c)( b - c) +( c + a)(c - a) +( a + b)(a - b)}{(c - a)(a - b)( b - c)} } \\ \\ \\ \rm \large \longrightarrow {x}^{ \frac{{b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} + {a}^{2} - {b}^{2} }{(c - a)(a - b)( b - c)} } \\ \\ \\ \rm \large \longrightarrow {x}^{ \frac{{b}^{2} - {b}^{2} + {c}^{2} - {c}^{2} + {a}^{2} - {a}^{2} }{(c - a)(a - b)( b - c)} } \\ \\ \\ \rm \large \longrightarrow {x}^{ \frac{0 }{(c - a)(a - b)( b - c)} } = {x}^{0} \\ \\ \\ \rm \large \longrightarrow {x}^{0} = 1$

Therefore, RHS = LHS.

Hence Proved