Question

Please reply this;
(x^b+c/c-a)^1/a+b*(x^c+a/a-b)^1/b+c*(x^a+b/x^b-c) ^1/c+a

Answer 1

Hi ,

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

I hope this helps you.

:)