Question

prove that (x^a/x^b)a+b (x^b/x^c) b+c (x^c/x^a) c+a =1 . It's hurry pls ans i Will mark as brainliest

Answer 1

Step-by-step explanation:

We need to prove the gives equation is unity that is 1

LHS=(xa/xb)^1/ab( xb /xc)^1/bc(xc/xa)1/ca

Using laws of exponents

= (xa/xb)1/ab( xb /xc)1/bc(xc/xa)1/ca

= x(a-b)/ab * x^(b-c)/bc * x^(c-a)/ca

= x[(a-b)/ab + (b-c)/bc + (c-a)/ca]

= x[c(a-b)/abc + a(b-c)/abc + b(c-a)/abc ]

= x { [c(a-b)+ a(b-c) + b(c-a) ]/abc }

= x ( ac – bc + ab – ac + bc – ab ] /abc

= x 0/abc

= x0

= 1

= RHS

Hence proved

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Answer 2

Answer:

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${( \frac{ {x}^{a} }{ {x}^{b} }) }^{ \: a + b \: } \times {( \frac{ {x}^{b} }{ {x}^{c} }) }^{ \: b + c \: } \times {( \frac{ {x}^{c} }{ {x}^{a} }) }^{ \: c+ a \: }$

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

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

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

$= {x}^{0}$

$= 1$