Question

(x^(a+b)) ^2 ((x^(b+c))^2 (x^(c+a))^2 / [(x^a) (x^b) (x^c) ]
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Answer 1

Answer:

(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

sorry if its wrong i tried my best