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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.
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