Simplify:$ \left(\frac{x^{a+b}}{x^{c}}\right)^{a-b}\left(\frac{x^{b+c}}{x^{a}}\right)^{b-c}\left(\frac{x^{c+a}}{x^{b}}\right)^{c-a} $

Given:

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

To do: 

We have to simplify the given expression.

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,

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

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

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

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

$=x^{0}$

$=1$

Hence, $(\frac{x^{a+b}}{x^{c}})^{a-b}(\frac{x^{b+c}}{x^{a}})^{b-c}(\frac{x^{c+a}}{x^{b}})^{c-a}=1$.

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Updated on: 10-Oct-2022

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