Question

(x^a+b)^a-b.(x^b+a)^b-c.(x^c+a)^c-a​

Answer 1

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