Show that:

\( \left(\frac{3^{a}}{3^{b}}\right)^{a+b}\left(\frac{3^{b}}{3^{c}}\right)^{b+c}\left(\frac{3^{c}}{3^{a}}\right)^{c+a}=1 \)

To do: 

We have to show that \( \left(\frac{3^{a}}{3^{b}}\right)^{a+b}\left(\frac{3^{b}}{3^{c}}\right)^{b+c}\left(\frac{3^{c}}{3^{a}}\right)^{c+a}=1 \).

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

LHS $=(\frac{3^{a}}{3^{b}})^{a+b}(\frac{3^{b}}{3^{c}})^{b+c}(\frac{3^{c}}{3^{a}})^{c+a}$

$=(3^{a-b})^{a+b} \times(3^{b-c})^{b+c} \times(3^{c-a})^{c+a}$

$=3^{a^{2}-b^{2}} \times 3^{b^{2}-c^{2}} \times 3^{c^{2}-a^{2}}$

$=3^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}$

$=3^{0}$

$=1$

$=$ RHS

Hence proved.          

Tutorialspoint

Updated on: 10-Oct-2022

40 Views

Kickstart Your Career

Get certified by completing the course

Get Started