Question

Prove ${\left(\frac{2^a}{2^b}\right)}^{a+b}\times {\left(\frac{2^b}{2^c}\right)}^{b+c}\times {\left(\frac{2^c}{2^a}\right)}^{c+a}=1$

Answer 1

$(\frac{2^a}{2^b}{)}^{a+b}\times (\frac{2^b}{2^c}{)}^{b+c}\times (\frac{2^c}{2^a}{)}^{c+a}=(2^{a-b}{)}^{a+b}+(2^{b-c}{)}^{b+c}+(2^{c-a}{)}^{c+a}$ (since, $\frac{a^m}{a^n}=a^{m-n}$)

$=2^{(a-b)(a+b)}\times 2^{(b-c)(b+c)}\times 2^{(c-a)(c+a)}$ (since, $\left(a^m\right)n=a^{mn}$)

$=2^{(a^2-b^2)}\times 2^{(b^2-c^2)}\times 2^{(c^2-a^2)}$

$=2^{a^2-b^2+b^2-c^2+c^2-a^2}$ (since, $a^m\times a^n=a^{m+n}$)

$=2^0=1$