Show that:$ \left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}=1 $

To do: 

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

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

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

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

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

$=x^{0}$

$=1$

$=$ RHS

Hence proved.