Prove that:$ \left(\frac{x^{a}}{x^{b}}\right)^{c} \times\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b}=1 $

Given:

\( \left(\frac{x^{a}}{x^{b}}\right)^{c} \times\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b}=1 \)

To do:

We have to prove that \( \left(\frac{x^{a}}{x^{b}}\right)^{c} \times\left(\frac{x^{b}}{x^{c}}\right)^{a} \times\left(\frac{x^{c}}{x^{a}}\right)^{b}=1 \).

Solution:

We know that,

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

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

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

$a^{0}=1$  

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

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

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

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

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

$=x^{0}$

$=1$

$=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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