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Show that:
\( \left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1 \)
To do:
We have to show that \( \left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1 \)
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 $=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$
$=x^{\frac{1}{(a-b)} \times \frac{1}{(a-c)}} \times x^{\frac{1}{b-c} \times \frac{1}{b-a}} \times x^{\frac{1}{c-a} \times \frac{1}{c-b}}$
$=x^{\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}}$
$=x^{\frac{-1}{(a-b)(c-a)}+\frac{-1}{(b-c)(a-b)}+\frac{-1}{(c-a)(b-c)}}$
$=x^{\frac{-b+c-c+a-a+b}{(a-b)(b-c)(c-a)}}$
$=x^{\frac{0}{(a-b)(b-c)(c-a)}}$
$=x^{0}$
$=1$
$=$ RHS
Hence proved.