Question

Prove that

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

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

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

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

Answer 1

(i) $\sqrt{x^{-1}y}·\sqrt{y^{-1}z}·\sqrt{z^{-1}x}=1$
$$\begin{gathered} \mathrm{LHS}=\sqrt{x^{-1}y}·\sqrt{y^{-1}z}·\sqrt{z^{-1}x} \\ ={\left(x^{-1}y\right)}^{\frac{1}{2}}·{\left(y^{-1}z\right)}^{\frac{1}{2}}·{\left(z^{-1}x\right)}^{\frac{1}{2}} \\ =\left(x^{-\frac{1}{2}}{y}^{\frac{1}{2}}\right)·\left(y^{-\frac{1}{2}}{z}^{\frac{1}{2}}\right)·\left(z^{-\frac{1}{2}}{x}^{\frac{1}{2}}\right) \\ =x^{-\frac{1}{2}+\frac{1}{2}}{y}^{\frac{1}{2}-\frac{1}{2}}{z}^{\frac{1}{2}-\frac{1}{2}} \\ =x^0{y}^0{z}^0 \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

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

(ii) ${\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$
$$\begin{gathered} \mathrm{LHS}={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{-\frac{1}{c-b}}\right)}^{\frac{1}{b-a}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{-\frac{1}{b-a}}\right)}^{\frac{1}{c-b}}·{\left(x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{-\frac{1}{b-a}}.x^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{1}{c-a}-\frac{1}{b-a}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{b-a-c+a}{\left(c-a\right)\left(b-a\right)}}\right)}^{\frac{1}{c-b}} \\ ={\left(x^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left(x^{\frac{b-c}{\left(c-a\right)\left(b-a\right)}}\right)}^{-\frac{1}{b-c}} \\ =\left(x^{\frac{1}{\left(a-b\right)\left(a-c\right)}}\right)·\left(x^{\frac{-1}{\left(c-a\right)\left(b-a\right)}}\right) \\ =\left(x^{\frac{1}{\left(b-a\right)\left(c-a\right)}}\right)·\left(x^{-\frac{1}{\left(c-a\right)\left(b-a\right)}}\right) \\ =\left(x^{\frac{1}{\left(b-a\right)\left(c-a\right)}-\frac{1}{\left(c-a\right)\left(b-a\right)}}\right) \\ =x^0 \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

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

(iii) $\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}÷{\left(\frac{x^b}{x^a}\right)}^c=1$
$$\begin{gathered} \mathrm{LHS}=\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}÷{\left(\frac{x^b}{x^a}\right)}^c \\ =\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}\times {\left(\frac{x^a}{x^b}\right)}^c \\ =\frac{x^{a\left(b-c\right)}}{x^{b\left(a-c\right)}}\times \frac{x^{ac}}{x^{bc}} \\ =\frac{x^{ab-ac}}{x^{ba-bc}}\times \frac{x^{ac}}{x^{bc}} \\ =x^{ab-ac-ba+bc}.x^{ac-bc} \\ =x^{-ac+bc}.x^{ac-bc} \\ =x^{-ac+bc+ac-bc} \\ =x^0 \\ =1 \\ =\mathrm{RHS} \end{gathered}$$

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

(iv) $\frac{{\left(x^{a+b}\right)}^2{\left(x^{b+c}\right)}^2{\left(x^{c+a}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=1$
$$\begin{gathered} \mathrm{LHS}=\frac{{\left(x^{a+b}\right)}^2{\left(x^{b+c}\right)}^2{\left(x^{c+a}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4} \\ =\frac{\left(x^{2a+2b}\right)\left(x^{2b+2c}\right)\left(x^{2c+2a}\right)}{\left(x^{4a}{x}^{4b}{x}^{4c}\right)} \\ =\frac{x^{2a+2b+2b+2c+2c+2a}}{x^{4a+4b+4c}} \\ =\frac{x^{4a+4b+4c}}{x^{4a+4b+4c}} \\ =1 \\ =\mathrm{RHS} \end{gathered}$$
Hence, $\frac{{\left(x^{a+b}\right)}^2{\left(x^{b+c}\right)}^2{\left(x^{c+a}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=1$.