Question

Show that: $\sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1.$

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Answer 1

Here we need to Prove that ,

$\bf\implies \sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1.$

Lets start with LHS , which is ,

$\bf\implies LHS = \sqrt[ab]{ \dfrac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \dfrac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \dfrac{ {x}^{c} }{ {x}^{a} } }$

We know that ,

$\bf\implies \sqrt[n]{k^m }= k^{\frac{m}{n}}$

• So that ,

$\bf\implies \sqrt[ab]{x^{a-b}} . \sqrt[bc]{x^{b-c}} .\sqrt[ac]{x^{c-a}} \\\\\bf\implies x^{\frac{a-b}{ab}} . x ^{\frac{b-c}{bc}} .x^{\frac{c-a}{ca}} \\\\\bf\implies x^{\frac{ ac - bc + ab - ac + bc - ba }{abc } }\\\\\bf\implies x^{\frac{0}{abc}} \\\\\bf\implies x^0 \\\\\bf\implies 1 = RHS$

Hence ,

$\implies \boxed{\bf \red{ LHS = RHS }}$

Answer 2

$\large\underline{\sf{Solution-}}$

Consider LHS:

$\sf= \sqrt[\sf ab]{\dfrac{\sf x^{a}}{\sf x^{b}}}\cdot \sqrt[\sf bc]{\dfrac{\sf x^{b}}{\sf x^{c}}}\cdot \sqrt[\sf ac]{\dfrac{\sf x^{c}}{\sf x^{a}}}$

As we know that:

$\sf\red\leadsto\dfrac{x^{a}}{x^{b}}=x^{a-b}$

We get:

$\sf= \sqrt[\sf ab]{\sf x^{a-b}}\cdot \sqrt[\sf bc]{\sf x^{b-c}}\cdot \sqrt[\sf ac]{\sf x^{c-a}}$

We can also write it as:

$\sf= x^{\dfrac{a-b}{ab}}\cdot x^{\dfrac{b-c}{bc}}\cdot x^{\dfrac{c-a}{ac}}$

We know that:

$\sf \red\leadsto x^{a}\cdot x^{b}=x^{a+b}$

We get:

$\sf= x^{\bigg(\dfrac{a-b}{ab}+\dfrac{b-c}{bc}+\dfrac{c-a}{ac}\bigg)}$

$\sf= x^{\bigg(\dfrac{c(a-b)+a(b-c)+b(c-a)}{abc}\bigg)}$

$\sf= x^{\bigg(\dfrac{ac-bc+ab-ac+bc-ab}{abc}\bigg)}$

$\sf= x^{\bigg(\dfrac{0}{abc}\bigg)}$

$\sf= x^{0}$

$\sf=1$

$\textsf{= RHS}$