Question

show (x⁹/x^b) 1/ab (x^bx^c)1/bc (x^c/x^a)1/ac=1 ​

Answer 1

Answer:Step-by-step explanation

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

:

Answer 2

Answer:

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