Prove thatCl.abrc

1.	Prove thatCl.abrc
Answer» we know that 1 ) x^m / x^n = x^(m - n) 2 ) (x^m)^n = x^mn 3 ) x^0 = 1 4) x^m * x^n = x^(m+n) Now, LHS=(x^a/x^b)^1/ab( x^b /x^c)^1/bc(x^c/x^a)^1/ca = (x^a-b)^1/ab(x^b-c)^1/bc(x^c-a)^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 = x^0 = 1 = RHS Hence proved

Answer»

we know that

1 ) x^m / x^n = x^(m - n)

2 ) (x^m)^n = x^mn

3 ) x^0 = 1

4) x^m * x^n = x^(m+n)

Now, LHS=(x^a/x^b)^1/ab( x^b /x^c)^1/bc(x^c/x^a)^1/ca

= (x^a-b)^1/ab(x^b-c)^1/bc(x^c-a)^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

= x^0

= 1

= RHS

Hence proved

Discussion