Question

Show that:
(x^a/x^b)^1/ab (x^b/x^c)^1/bc (x^c/x^a)^1/ca =1

Answer 1

$[(\frac{x^a}{x^b})^{\frac{1}{ab}}][(\frac{x^b}{x^c})^{\frac{1}{bc}}][(\frac{x^c}{x^a})^{\frac{1}{ca}}] \\ \\ [ (x^{a-b})^{\frac{1}{ab}}][(x^{b-c})^{\frac{1}{bc}}][(x^{c-a})^{\frac{1}{ca}}] \ \ \ \ \ \ \ [\frac{x^m}{x^n}=x^{m-n}] \\ \\ [ x^{\frac{a-b}{ab}}][x^{\frac{b-c}{bc}}][x^{\frac{c-a}{ca}}] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [(x^m)^n=x^{mn}]$

$[x^{\frac{1}{b}-\frac{1}{a}}][x^{\frac{1}{c}-\frac{1}{b}}][x^{\frac{1}{a}-\frac{1}{c}}] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\frac{m-n}{mn}=\frac{1}{n}-\frac{1}{m}] \\ \\ x^{(\frac{1}{b}-\frac{1}{a})+(\frac{1}{c}-\frac{1}{b})+(\frac{1}{a}-\frac{1}{c})} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [x^m \times x^n=x^{m+n}] \\ \\ x^{\frac{1}{b}-\frac{1}{a}+\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}} \\ \\ x^{\frac{1}{b}-\frac{1}{b}+\frac{1}{c}-\frac{1}{c}+\frac{1}{a}-\frac{1}{a}}$

$x^0 \\ \\ \bold{1}$

$Hence it's proved!!! \\ \\ \\ Thank you. Have a nice day. :-) \\ \\ \\ \\ \\ \#adithyasajeevan$

Answer 2

Formulas required :

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

x^m . x^n = x^( m + n )

PROOF

( 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 )/ac

⇒ x^[ ( a - b )/ ab + ( b - c ) / bc + ( c - a ) / ac ]

⇒ x^[ ( ac - bc + ab - ac + bc - ab ) / abc ]

⇒ x^[ 0 / abc ]

⇒ x^0

⇒ 1