# 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.

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