Question

please explain me.how to solve it (i) (xa/xb)1/ab (xb/xc)1/bc (xc/xa)1/ca  = 1

L.H.S. = (xa/xb)1/ab (xb/xc)1/bc (xc/xa)1/ca

= (x a-b)1/b (xb-c)1/bc (xc-a)1/ca

= x a-b/ab xb-c/bc x c-a/ca

{(xa)b = xab}

= x a-b/ab + b-c/bc +c-a/ca

= x (ac – bc + ab – ac + bc – ab)/abc

= x0 = 1 = R.H.S  (∵XO = 1)​

Answer 1

Step-by-step explanation:

Given: (xa/xb)^1/ab( xb /xc)^1/bc(xc/xa)1/ca

We need to prove the gives equation is unity that si 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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