Question

If x is a positive real number and the exponents are rational numbers, show that :

${\left[\frac{x^a}{x^b}\right]}^{a+b-c}{\left[\frac{x^b}{x^c}\right]}^{b+c-a}{\left[\frac{x^c}{x^a}\right]}^{c+a-b}=1$

Answer 1

We have to prove that, ${\left[\frac{x^a}{x^b}\right]}^{a+b-c}{\left[\frac{x^b}{x^c}\right]}^{b+c-a}{\left[\frac{x^c}{x^a}\right]}^{c+a-b}=1$

Let $a={\left[\frac{x^a}{x^b}\right]}^{a+b-c}{\left[\frac{x^b}{x^c}\right]}^{b+c-a}{\left[\frac{x^c}{x^a}\right]}^{c+a-b}$

$=\left(x^{a-b}{)}^{a+b-c}\right(x^{b-c}{)}^{b+c-a}(x^{c-a}{)}^{c+a-b}$

$=x^{(a-b)(a+b-c)}\times x^{(b-c)(b+c-a)}\times x^{(c-a)(c+a-b)}$

$=x^{(a-b)(a+b)-c(a-b)}\times x^{(b-c)(b+c)-a(b-c)}\times x^{(c-a)(c+a)-b(c-a)}$

$=x^{a^2-b^2-ca+bc}\times x^{b^2-c^2-ab+ac}\times x^{c^2-a^2-bc+ba}$

$a=x^0=1$

$\therefore {\left[\frac{x^a}{x^b}\right]}^{a+b-c}{\left[\frac{x^b}{x^c}\right]}^{b+c-a}{\left[\frac{x^c}{x^a}\right]}^{c+a-b}=1$