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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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