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Show that:$ \left[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\} \p\left\{\frac{x^{b(b-a)}}{x^{b(b+a)}}\right\}\right]^{a+b}=1 $
To do:
We have to show that \( \left[\left\{\frac{x^{a(a-b)}}{x^{a(a+b)}}\right\} \div\left\{\frac{x^{b(b-a)}}{x^{b(b+a)}}\right\}\right]^{a+b}=1 \)
Solution:
We know that,
$(a^{m})^{n}=a^{m n}$
$a^{m} \times a^{n}=a^{m+n}$
$a^{m} \div a^{n}=a^{m-n}$
$a^{0}=1$
Therefore,
LHS $=[{\frac{x^{a(a-b)}}{x^{a(a+b)}}} \div{\frac{x^{b(b-a)}}{x^{b(b+a)}}}]^{a+b}$
$=[\frac{x^{a^2-ab}}{x^{a^2+ab}}\div\frac{x^{b^2-ab}}{x^{b^2+ab}}]^{a+b}$
$=[x^{a^2-ab-a^2-ab} \div x^{b^2-ab-b^2-ab}]^{a+b}$
$=[x^{-2ab} \div x^{-2ab}]^{a+b}$
$=[x^{-2ab-(-2ab)}]^{a+b}$
$=[x^{-2ab+2ab}]^{a+b}$
$=(x^0)^{a+b}$
$=(1)^{a+b}$
$=1$
$=$ RHS
Hence proved.
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