((x^a)^b)^1/ab((x^b)^c)^1/bc((x^c)^a)^1/ca
Answers
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
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