Prove that

Solution:

(i) $\sqrt{x^{-1} y} \cdot \sqrt{y^{-1} z} \cdot \sqrt{z^{-1} x}=1$

LHS $=\sqrt{x^{-1} y} \cdot \sqrt{y^{-1} z} \cdot \sqrt{z^{-1} x}$

$=\left(x^{-1} y\right)^{\frac{1}{2}} \cdot\left(y^{-1} z\right)^{\frac{1}{2}} \cdot\left(z^{-1} x\right)^{\frac{1}{2}}$

$=\left(x^{-\frac{1}{2}} y^{\frac{1}{2}}\right) \cdot\left(y^{-\frac{1}{2}} z^{\frac{1}{2}}\right) \cdot\left(z^{-\frac{1}{2}} x^{\frac{1}{2}}\right)$

$=x^{-\frac{1}{2}+\frac{1}{2}} y^{\frac{1}{2}-\frac{1}{2}} z^{\frac{1}{2}-\frac{1}{2}}$

$=x^{0} y^{0} z^{0}$

$=1$

$=$ RHS

Hence, $\sqrt{x^{-1} y} \cdot \sqrt{y^{-1} z} \cdot \sqrt{z^{-1} x}=1$.

(ii) $\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}} \cdot\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1$

$\mathrm{LHS}=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}} \cdot\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$

$=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{-\frac{1}{c-b}}\right)^{\frac{1}{b-a}} \cdot\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$

$=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{-\frac{1}{b-a}}\right)^{\frac{1}{c-b}} \cdot\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$

$=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{-\frac{1}{b-a}} \cdot x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}$

$=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{\frac{1}{c-a}-\frac{1}{b-a}}\right)^{\frac{1}{c-b}}$

$=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{\frac{b-a-c+a}{(c-a)(b-a)}}\right)^{\frac{1}{c-b}}$

$=\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{\frac{b-c}{(c-a)(b-a)}}\right)^{-\frac{1}{b-c}}$

$=\left(x^{\frac{1}{(a-b)(a-c)}}\right) \cdot\left(x^{\frac{-1}{(c-a)(b-a)}}\right)$

$=\left(x^{\frac{1}{(b-a)(c-a)}}\right) \cdot\left(x^{-\frac{1}{(c-a)(b-a)}}\right)$

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

$=x^{0}$

$=1$

$=\mathrm{RHS}$

Hence, $\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}} \cdot\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1$.

(iii) $\frac{x^{a(b-c)}}{x^{b(a-c)}} \div\left(\frac{x^{b}}{x^{a}}\right)^{c}=1$

$\mathrm{LHS}=\frac{x^{a(b-c)}}{x^{b(a-c)}} \div\left(\frac{x^{b}}{x^{a}}\right)^{c}$

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

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

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

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

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

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

$=x^{0}$

$=1$

$=\mathrm{RHS}$

Hence, $\frac{x^{a(b-c)}}{x^{b(a-c)}} \div\left(\frac{x^{b}}{x^{a}}\right)^{c}=1$.

(iv) $\frac{\left(x^{a+b}\right)^{2}\left(x^{b+c}\right)^{2}\left(x^{c+a}\right)^{2}}{\left(x^{a} x^{b} x^{c}\right)^{4}}=1$

$\mathrm{LHS}=\frac{\left(x^{a+b}\right)^{2}\left(x^{b+c}\right)^{2}\left(x^{c+a}\right)^{2}}{\left(x^{a} x^{b} x^{c}\right)^{4}}$

$=\frac{\left(x^{2 a+2 b}\right)\left(x^{2 b+2 c}\right)\left(x^{2 c+2 a}\right)}{\left(x^{4 a} x^{4 b} x^{4 c}\right)}$

$=\frac{x^{2 a+2 b+2 b+2 c+2 c+2 a}}{x^{4 a+4 b+4 c}}$

$=\frac{x^{4 a+4 b+4 c}}{x^{4 a+4 b+4 c}}$

$=1$

$=$ RHS

Hence, $\frac{\left(x^{a+b}\right)^{2}\left(x^{b+c}\right)^{2}\left(x^{c+a}\right)^{2}}{\left(x^{a} x^{b} x^{c}\right)^{4}}=1$.